[QUOTE=3.14159;225036]A program that generates the Fermat pseudoprimes in a given range, that are pseudoprime to user-selected bases. (No need for a code snippet, I eventually remembered how to write it and saved it in the text document, along with the other functions.[/QUOTE]

Ah. Hard task!

[QUOTE=3.14159;225040]An interesting application error:
Whenever I input 25326001, it returns that it is a 7-SPRP, when it is in fact reported "composite" using base 7. I verified that it was an error on the app. Now I see that this is an unreliable app, as it returns results that are clearly wrong.[/QUOTE]

What script are you using to check? My isSPRP gives
[code]>isSPRP(25326001,7)
%1 = 0[/code]

[QUOTE=science_man_88;225037][TEX]24m= 6np+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6np-(p+7)\right m=px+c ? [/TEX]
[TEX]if(px+c==(4n-1)/3,(4\strike n -1)/3[/TEX]
for all remaining n that create primes print(2*n+3) these are Mersenne prime exponents

get this CRG ?[/QUOTE]

sorry should be (4^n-1)/3 lol
oh and there shouldn't be 2 n equations lol.

[QUOTE=science_man_88;225037][TEX]24m= 6np+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6np-(p+7)\right m=px+c ? [/TEX]
[TEX]if(px+c==(4n-1)/3,(4\strike n -1)/3[/TEX]
for all remaining n that create primes print(2*n+3) these are Mersenne prime exponents

get this CRG ?[/QUOTE]

Hmm. Unlike the diagrams, I feel like I'm close to understanding something. The first two lines are clearly related to your repeating theme,
[TEX]24m+7=p(6n\pm1)[/TEX].

The lines themselves seem to be statements of the division algorithm. The first says that if 24m+1 is of the form p(6n+1), then p(6n+1)/24 = px + c, that is, p(6n+1) leaves a remainder of c when divided by p.

But wait, it doesn't matter which of the first two lines I choose -- px + c is always m. So the first two lines say that 24m+7 is composite (assuming that n > 0 and p > 1) and the third line says that m is (4n-1)/3.

OK, so m = (4n-1)/3 means 3m+1 = 4n, so the third line is saying that [TEX]m\equiv1\pmod4[/TEX]. So let m = 4k+1, then 3(4k+1)+1=4n, so that n = 3k + 1.

The original composite is then 24m+7 = 24(4k + 1) + 7 = 96k + 31 and the Mersenne non-exponent is 2n+3 = 2(3k+1)+1=6k+3.

And indeed yes, this is correct: if k is a positive integer, then 6k+3 is not a Mersenne exponent. But this won't remove all bad exponents, of course.

[QUOTE=science_man_88;225047]sorry should be (4^n-1)/3 lol
oh and there shouldn't be 2 n equations lol.[/QUOTE]

Pity, since I just analyzed it without.

Re-write it with those corrections and I'll see what I can do.

[TEX]24m= 6tp+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6tp-(p+7)\right m=px+c ? [/TEX]
[TEX]if(px+c==(4^n-1)/3,(4^\strike n -1)/3[/TEX]
for all remaining n [B]that create primes[/B] print(2*n+3) these are Mersenne prime exponents.

[QUOTE=science_man_88;225054][TEX]24m= 6tp+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6tp-(p+7)\right m=px+c ? [/TEX][/QUOTE]

If 24m + 7 is composite...

[QUOTE=science_man_88;225054][TEX]if(px+c==(4^n-1)/3,(4^\strike n -1)/3[/TEX][/QUOTE]

What does the [TEX],(4^\strike n -1)/3[/TEX] mean?

[QUOTE=science_man_88;225054]for all remaining n [B]that create primes[/B] print(2*n+3) these are Mersenne prime exponents.[/QUOTE]

For all n such that *what* is a prime? 2n + 3?

[QUOTE=CRGreathouse;225055]If 24m + 7 is composite...



What does the [TEX],(4^ \strike n -1)/3[/TEX] mean?



For all n such that *what* is a prime?[/QUOTE]

(4^n-1)/3 is the equation for A002450

that's what it was supposed to be then [TEX]\strike n[/TEX] (indexes) that are caught by px+c

what indexes are left over that give prime = 2*index+3 those should if my idea is flawless (never going to happen i'm guessing) be the indexes z such that [TEX]2^z-1[/TEX] are prime and in this case = Mersenne primes.

[QUOTE=science_man_88;225057](4^n-1)/3 is the equation for A002450

that's what it was supposed to be then [TEX]\strike n[/TEX] (indexes) that are caught by px+c

what indexes are left over that give prime = 2*index+3 those should if my idea is flawless (never going to happen i'm guessing) be the indexes z such that [TEX]2^z-1[/TEX] are prime and in this case = Mersenne primes.[/QUOTE]

Not following. I understand equations like
[TEX]px+c=(4^n-1)/3[/TEX]
but
[TEX]px+c=(4^n-1)/3,(4^\strike n -1)/3[/TEX]
doesn't mean anything to me.

[QUOTE=CRGreathouse;225059]Not following. I understand equations like
[TEX]px+c=(4^n-1)/3[/TEX]
but
[TEX]px+c=(4^n-1)/3,(4^ \strike n -1)/3[/TEX]
doesn't mean anything to me.[/QUOTE]

I was showing striking out the n in the first part with the second part for some reason it prints (4n-1)/3 even though i want n as a striked out exponent and I thought i coded as so.

[QUOTE=science_man_88;225061]I was striking out the n in the first part with the second for some reason it prints (4n-1)/3 even though i want n as a striked out exponent and I though i coded as so.[/QUOTE]

Uh...

Apparently you want do display something, I'm not sure exactly what (the n or the whole part with the n?), with a strike-through. But I don't know what a strike-through would mean in mathematics anyway, so I don't know what the equation is supposed to be.

Similarly, what does the comma mean?

[QUOTE=CRGreathouse;225062]Uh...

Apparently you want do display something, I'm not sure exactly what (the n or the whole part with the n?), with a strike-through. But I don't know what a strike-through would mean in mathematics anyway, so I don't know what the equation is supposed to be.

Similarly, what does the comma mean?[/QUOTE]

I was using the comma to separate the 2 parts to the last which is if the first equation proved true strike out the exponent n as an index that could give a Mersenne prime.

by the way you're lucky I even got tex tags working as I didn't know the syntax until I looked it up.

...and the last equation the one we are having trouble with comes from px+c being in A002450.

Assumptions: m, t, p, n are positive integers, p > 1.

[QUOTE=science_man_88;225054][TEX]24m= 6tp+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6tp-(p+7)\right m=px+c ? [/TEX][/QUOTE]

So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.

[QUOTE=science_man_88;225054][TEX]if(px+c==(4^n-1)/3,(4^\strike n -1)/3[/TEX][/QUOTE]

[QUOTE=science_man_88;225064]I was using the comma to separate the 2 parts to the last which is if the first equation proved true strike out the exponent n as an index that could give a Mersenne prime.[/QUOTE]

So if m is of the form (4^n-1)/3, I think you're you're saying one of these:
(a) n is not a Mersenne exponent (by which I mean that n isn't in A000043)
(b) (4n-1)/3 is not a Mersenne exponent
(c) (4^n-1)/3 is not a Mersenne exponent
(d) 2n+3 is not a Mersenne exponent

Could you clear this up?

[QUOTE=CRGreathouse;225067]Assumptions: m, t, p, n are positive integers, p > 1.



So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.





So if m is of the form (4^n-1)/3, I think you're you're saying one of these:
(a) n is not a Mersenne exponent (by which I mean that n isn't in A000043)
(b) (4n-1)/3 is not a Mersenne exponent
(c) (4^n-1)/3 is not a Mersenne exponent
(d) 2n+3 is not a Mersenne exponent

Could you clear this up?[/QUOTE]

d is pretty much on the button if you say 2n+3 must give a prime though then that seems you leaving only what is in A121290 to start with.

now the hard part speeding it up lol. I think including the z I used for [TEX]2^z-1[/TEX] that's 7 variables to deal with.

Assumptions: m, t, p, n are positive integers, p > 1.

[QUOTE=science_man_88;225054][TEX]24m= 6tp+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6tp-(p+7)\right m=px+c ? [/TEX][/QUOTE]

So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.

[QUOTE=science_man_88;225054][TEX]if(px+c==(4^n-1)/3,(4^\strike n -1)/3[/TEX][/QUOTE]

So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

[QUOTE=CRGreathouse;225074]Assumptions: m, t, p, n are positive integers, p > 1.



So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.



So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.[/QUOTE]


something tells me you are getting at a trivial answer ?

[QUOTE=CRGreathouse]What script are you using to check? My isSPRP gives
[/QUOTE]

Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.

[QUOTE=CRGreathouse]So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.
[/QUOTE]

Those are all trivially correct. There is no grand discovery here.

[QUOTE=3.14159;225082]Those are all trivially correct. There is no grand discovery here.[/QUOTE]

The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".

[QUOTE=3.14159;225081]Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.[/QUOTE]

Really? Interesting.

[QUOTE=CRGreathouse;225074]That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.[/QUOTE]

But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!

[QUOTE=CRGreathouse]The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".
[/QUOTE]

Wow. That was all that it was? A tautology/circular argument? :lol:!

[QUOTE=CRGreathouse]Really? Interesting.
[/QUOTE]

You can get it for yourself [URL="http://www.naturalnumbers.org/"]here[/URL].

I suspect it of being a kook site, but, seeing as their applets work correctly to some extent, I am undecided on that matter.

Try testing it out quickly (Note: Small integers only, please), and if you can catch 25326001 as an error, please note it!

[QUOTE=kar_bon]But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!
[/QUOTE]

Karsten, I found that you were right about the differences in testing times based on the k-values used.

[QUOTE=kar_bon;225085]But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime![/QUOTE]

The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.

Or rather, it was a restricted case of this: if n > 3 is an odd number such that 2^n - 1 is prime, then 2^n - 1 is prime.

[QUOTE=CRGreathouse]The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.
[/QUOTE]

Call the press! We're going to be filthy rich!

Amirite?

[SPOILER]Bullshit aside.. [/SPOILER]

CRG, have you managed to do some testing on 25326001 using that app? Does it say that it's a 7-SPRP?

[QUOTE=3.14159;225089]Wow. That was all that it was? A tautology/circular argument? :lol:![/QUOTE]

That's what I expected when it was brought up -- and I imagine I said as much on this thread (though I'm not going to go digging back through to look for it). But there's a

[QUOTE=3.14159;225089]I suspect it of being a kook site, but, seeing as their applets work correctly to some extent, I am undecided on that matter.[/QUOTE]

They're trying pretty hard to look like kooks, but perhaps they're not. I'll have a look at their code. VB... [i]shudder[/i].

[QUOTE=3.14159;225092]Call the press! We're going to be filthy rich!

Amirite?[/QUOTE]

I expect to see the royalty checks coming for sm88 soon enough. After that, fast cars and women lined up 'round the block. But I'll be happy to bask in the reflected glory.

All kidding aside, though, the discovery did pan out -- it just didn't end up giving additional insight or faster techniques. What's more, sm88 is better able to express his mathematical ideas. Surely you find post #297 easier to understand than post #28? So there has been serious progress. And of course he also now knows how to use Pari.

[QUOTE=3.14159;225092]CRG, have you managed to do some testing on 25326001 using that app? Does it say that it's a 7-SPRP?[/QUOTE]

I downloaded the source code and I'm looking it over. It's interesting -- almost all of the math wasn't written by the naturalnumbers guy, but rather by a David Ireland who actually knows some math (though he's clearly not a mathematician, and his code is not fast). The difference between the different parts of the code is crazy.

But I haven't actually run the program yet.

[QUOTE=CRGreathouse]That's what I expected when it was brought up -- and I imagine I said as much on this thread (though I'm not going to go digging back through to look for it). But there's a
[/QUOTE]

It looks as if this sentence is unfinished. If so, what did you intend to post?

[QUOTE=CRGreathouse]I expect to see the royalty checks coming for sm88 soon enough. After that, fast cars and women lined up 'round the block. But I'll be happy to bask in the reflected glory.
[/QUOTE]

We'll be celebrities! We'll be known as math geniuses everywhere! We'll become household names!

[QUOTE=CRGreathouse]All kidding aside, though, the discovery did pan out -- it just didn't end up giving additional insight or faster techniques. What's more, sm88 is better able to express his mathematical ideas. Surely you find post #297 easier to understand than post #28? So there has been serious progress. And of course he also now knows how to use Pari.
[/QUOTE]

Admittedly, he did make some progress in expressing his ideas. I was surprised you were able to understand what he meant most of the time.

[QUOTE=CRGreathouse]I downloaded the source code and I'm looking it over. It's interesting -- almost all of the math wasn't written by the naturalnumbers guy, but rather by a David Ireland who actually knows some math (though he's clearly not a mathematician, and his code is not fast).[/QUOTE]

In the site, Ross claims:

[QUOTE=Michael M. Ross](Modular exponentiation code provided by DI Management Cryptography Software.)[/QUOTE]

[QUOTE=CRGreathouse]They're trying pretty hard to look like kooks, but perhaps they're not. I'll have a look at their code. VB... shudder.
[/QUOTE]

Just read their article on Fermat's factoring method: They claim they made improvements to the method, and you can view those [URL="http://www.naturalnumbers.org/fixfermatfact.html"]here[/URL].

[QUOTE=3.14159;225104]It looks as if this sentence is unfinished. If so, what did you intend to post?[/QUOTE]

I did get cut off there (accidental shift-end, I imagine), but I continued the thoughts in the next post.

[QUOTE=3.14159;225104]Just read their article on Fermat's factoring method: They claim they made improvements to the method[/QUOTE]

I saw that too. The 'improvements' look like the usual way it's used, and Fermat's method is basically worthless with or without improvements. (It does lead to better methods, eventually up to the NFS -- but on its own it's far too slow.)

[QUOTE=CRGreathouse]I saw that too. The 'improvements' look like the usual way it's used, and Fermat's method is basically worthless with or without improvements. (It does lead to better methods, eventually up to the NFS -- but on its own it's far too slow.)
[/QUOTE]

It is useless for any numbers larger than about 10-12 digits. Trial division is better than it.

[QUOTE= Kook]*OK, I acknowledge that trial div still has the edge for this baby example. Obviously, try something bigger for a fair test, say 9 or 10 digits. Remember that we're interested in the difficult ones - those with only 2 prime factors that are similar in size to the square root. A systematic survey by magnitude will be coming soon.
[/QUOTE]

10 digits? TD can find such a number's factors instantly. One of his own apps takes about a tenth of a second to find a p5. Fermat's is useless.

Another thing to ponder: Can QS be done, manually, w/no compy help, given enough time and knowledge of how it works (And maybe a large enough sheet of paper and a pen with plenty of ink)?

That would make decent headlines:

[QUOTE=Imaginary Headlines:]
Attention: A world record has been broken! Mr. Joe Smith has successfully split the c37 number 3868201127095457574466947223184535677 into its prime factors, with no computer assistance!:

[B]92734364657447299 * 41712704253533812223 ! [/B]

It took him 45 days to accomplish this. Mr. Joe Smith is looking forward to his $1500 reward.

[/QUOTE]

P.S: And no, I do not mean "41712704253533812223 factorial"

So, all circular stuffs aside:

Still conducting the prime searches for a ≈119000-digit prime and a 257920-digit prime. (Using bases 2 and 798336)

What is the knowledge on A165223 I know it's Products of 2 successive Mersenne primes. I want to know about something known before I post.

I know nothing special about that sequence. Essentially the only way to compute terms is to find all the Mersenne primes in a given range.

I can say things based on my knowledge of Mersenne primes, though. For example, for n > 2, a(n+1) > 64a(n).

I find all the terms except the first have a digital root of 4 or 1 just like the Mersenne primes >7 if we could confirm something like a pattern (I think starting at the 4th term they may take on the opposite to the series for the Mersenne primes>31) if this was ever proven true it could let us look 4 Mersenne primes in the future(that's why I'm doubtful).

[QUOTE=science_man_88]I find all the terms except the first have a digital root of 4 or 1 just like the Mersenne primes >7 if we could confirm something like a pattern (I think starting at the 4th term they may take on the opposite to the series for the Mersenne primes>31) if this was ever proven true it could let us look 4 Mersenne primes in the future(that's why I'm doubtful).
[/QUOTE]

The law of small numbers, at work again. Unless you can definitively prove this, it is nothing more than a guess.

[QUOTE=science_man_88;225135]I find all the terms except the first have a digital root of 4 or 1 just like the Mersenne primes >7 if we could confirm something like a pattern (I think starting at the 4th term they may take on the opposite to the series for the Mersenne primes>31) if this was ever proven true it could let us look 4 Mersenne primes in the future(that's why I'm doubtful).[/QUOTE]

Digital roots are essentially the number mod 9. Mersenne numbers other than 3 are of the form 2^(2k+1) - 1 = 4^k * 2 - 1, which is 1 mod 3. Thus Mersenne numbers other than 3 are 1, 4, or 7 mod 9. Further, the numbers are 7 mod 9 only when the exponent is a multiple of 3, which for Mersenne primes happens only with M_3 = 7. So the product of two Mersenne primes, both greater than 7, will be 1*1, 1*4, 4*1, or 4*4 mod 9, that is, will have digital root 1, 4, or 7.

You only gave the first two possibilities, but the third happens with
(2[SUP]89[/SUP] - 1)(2[SUP]107[/SUP] - 1) = 100433627766186892221372630609062766858404681029709092356097
and presumably occurs infinitely often.

[QUOTE=3.14159;225136]The law of small numbers, at work again. Unless you can definitively prove this, it is nothing more than a guess.[/QUOTE]

It seems to show that science_man_88's intuition is pretty sharp [i]vis-a-vis[/i] the strong law of small numbers, actually. :smile: Although he did miss one case, I guess because it wasn't listed in [url=http://oeis.org/classic/A165223]A165223[/url].

[QUOTE=CRGreathouse]It seems to show that science_man_88's intuition is pretty sharp [I]vis-a-vis[/I]the strong law of small numbers, actually. :smile: Although he did miss one case, I guess because it wasn't listed in A165223.
[/QUOTE]

Remember: Just because it works for three examples, doesn't mean it works forever.

[QUOTE=3.14159;225139]Remember: Just because it works for three examples, doesn't mean it works forever.[/QUOTE]

funny I checked every one it the sequence.

Also: Are there any sieving programs for n * k! + 1 ?

Nevermind: PARI does the work here. I'll set up a program which can act as an amateur sieve:

kfacsieve(x, n, a, m) = {
for(p=x,n,
if((x*n+1)%p==0, ...

[QUOTE=3.14159;225141]Also: Are there any sieving programs for n * k! + 1 ?

Nevermind: PARI does the work here. I'll set up a program which can act as an amateur sieve:

kfacsieve(x, n, a, m) = {
for(p=x,n,
if((x*n+1)%p==0, ...[/QUOTE]

I got [CODE]for(n=1,100,for(k=1,100,if(isprime(n*k!+1),print(n","k))))[/CODE]

[B]Without[/B] using (isprime(x))

Also: @CRG: I tried this:

[code] d(a,n,x,m)=for(n=a,x,if(t(n*m!+1)!=n,print("trivial composite","factor is",t(n*m!+1)));if(t(n*m!+1)==n,print(n*m!+1)))[/code]

Where t(n) is trial division up to nextprime(10^6).

I get the error: "Too many parameters in user-defined function".

I don't see where the excess params are at: I introduced no new variables into the commands. Or is it treating t as a variable?

[QUOTE=3.14159;225141]Also: Are there any sieving programs for n * k! + 1 ?[/QUOTE]

What's fixed?

[QUOTE=CRGreathouse]What's fixed?
[/QUOTE]

k is fixed, n is variable.

Reason: Primeform too slow!

[QUOTE=3.14159;225149]k is fixed, n is variable.

Reason: Primeform too slow![/QUOTE]

I got [CODE] not a function: `t'.[/CODE]

posting it into pari.

Because [B][COLOR="Red"]you have not defined t yet. I have.[/COLOR][/B]

[QUOTE=3.14159;225145][code] d(a,n,x,m)=for(n=a,x,if(t(n*m!+1)!=n,print("trivial composite","factor is",t(n*m!+1)));if(t(n*m!+1)==n,print(n*m!+1)))[/code]

Where t(n) is trial division up to nextprime(10^6).

I get the error: "Too many parameters in user-defined function".

I don't see where the excess params are at: I introduced no new variables into the commands. Or is it treating t as a variable?[/QUOTE]

Can you show me your code for t?

Also, you're computing t(n*m!+1) twice, which you can avoid like so:
[code]d(a,n,x,m)={
for(n=a,x,
if(t(n*m!+1)!=n,
print("trivial composite","factor is",t(n*m!+1))
[COLOR="Red"]);
if(t(n*m!+1)==n,[/COLOR]
print(n*m!+1)
)
)
};[/code]
[code]d(a,n,x,m)={
for(n=a,x,
if(t(n*m!+1)!=n,
print("trivial composite","factor is",t(n*m!+1))
[COLOR="Red"],[/COLOR]
print(n*m!+1)
)
)
};[/code]
using the form
[code]if(condition, do_if_true, do_if_false)[/code]
rather than
[code]if(condition, do_if_true);if(!condition, do_if_false)[/code]

another thing I notice is if you do it for multiplying 3 it starts 3,1,4 as well could this be a pattern continuation ? I'll have to check more but if it every was proven to be the same for higher numbers of them at a time we could say what's the largest index that is 0 Mod prime? number then figure out the mod of the unknown one through that and say only check primes that give a number with digital root x.

[QUOTE=science_man_88]another thing I notice is if you do it for multiplying 3 it starts 3,1,4 as well could this be a pattern continuation ? I'll have to check more but if it every was proven to be the same for higher numbers of them at a time we could say what's the largest index that is 0 Mod prime? number then figure out the mod of the unknown one through that and say only check primes that give a number with digital root x.
[/QUOTE]

Pardon, but, could you explain a bit more adequately?

[QUOTE=science_man_88;225180]another thing I notice is if you do it for multiplying 3 it starts 3,1,4 as well could this be a pattern continuation ? I'll have to check more but if it every was proven to be the same for higher numbers of them at a time we could say what's the largest index that is 0 Mod prime? number then figure out the mod of the unknown one through that and say only check primes that give a number with digital root x.[/QUOTE]

Are you still talking about the digital roots of A165223? Here are the first few:
3, 1, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 4, 1, 1, 1, 1, 4, 4, 4, 7, 7, 7, 7, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 7, 7, 7, 4

look:

3*7*31 = 217*3 = 651 6+5+1 = 12 1+2=3
7*31*127 = 27,559 10+18 effectively = 28 2+8 = 10 1+0 = 1
31*127*8191 = 5+6+7+9+6+7 effectively = 40 4+0 = 4

starts 3,1,4

try 5...

3*7*31*127*8191 = 677207307 6+7+7+2+7+3+7 = 39 3+9 = 12 1+2 = 3

however until I use Pari i can't check any further.

[QUOTE=science_man_88;225184]however until I use Pari i can't check any further.[/QUOTE]

I gave the first 38 above. A few more terms could probably be calculated; I don't remember how far the Mersenne numbers have been verified.

Ah, here's a program to get the digital root of a Mersenne number. The digital root of a product is the product of the digital roots, so you can use this if you like.

[code]drMersenne(expo)={
lift(Mod(2,9)^expo-1)
};
A43=[2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917];
vector(#A43-1,i,(drMersenne(A43[i])*drMersenne(A43[i+1]))%9)[/code]

what I'm saying is for prime numbers so far it looks as though it will always start 3,1,4

if we can get a sequence like the one you have above maybe we can predict modulo of the next Mersenne prime if we could it would possible eliminate some checks.

[QUOTE=science_man_88;225187]if we can get a sequence like the one you have above maybe we can predict modulo of the next Mersenne prime if we could it would possible eliminate some checks.[/QUOTE]

I've already shown that the sequence takes on all possible values for a Mersenne number with exponent relatively prime to 6. So all we can determine from the sequence is that further Mersenne exponents won't be divisible by 2 or 3.

@CRG: I found this number with the following properties:

49152096000^1+1 is prime. (49152096001)
49152096000^2+1 is prime. (2415928541193216000001)

And: 49152096000^3+1 has n^1+1 as its smallest prime factor:

49152096000^3+1 = 49152096001 * 2415928541144063904001

Can you find any other number with these properties?

Oh, wait: This is easy:

3
5
2^3+1 = 9 = 3*3

my idea is this:

[CODE]#M %9[SUB]1[/SUB] %9[SUB]2[/SUB] %9[SUB]3[/SUB]
2 3 1 4 x
3 3 1 4
5 3 [/CODE]

care to use the data in the table to calculate x ?

[QUOTE=3.14159;225193]Oh, wait: This is easy:

3
5
2^3+1 = 9 = 3*3[/QUOTE]

See? You've learned my trick of finding the smallest possible example! :smile:

[QUOTE=3.14159;225193]@CRG: I found this number with the following properties:

49152096000^1+1 is prime. (49152096001)
49152096000^2+1 is prime. (2415928541193216000001)

And: 49152096000^3+1 has n^1+1 as its smallest prime factor:

49152096000^3+1 = 49152096001 * 2415928541144063904001 [/QUOTE]

[code]forprime(p=2,1e4,
k=p*(p-2)+2;
if(isprime(k)&isprime(k-p+1),
print(p," ",k," ",(p-1)^3+1)
)
)[/code]

There are 835 such primes below a million, 4779 below ten million, and 31057 below a hundred million. (I'm referring to the first prime only; the other prime and the composite are of course larger.)

both ^1+1 and ^2+1 seem to have the same digital root.

[QUOTE=science_man_88;225194]my idea is this:

[CODE]#M %9[SUB]1[/SUB] %9[SUB]2[/SUB] %9[SUB]3[/SUB]
2 3 1 4 x
3 3 1 4
5 3 [/CODE]

care to use the data in the table to calculate x ?[/QUOTE]

I don't understand the table. What's x? What's #M? What do the center columns mean?

[QUOTE=science_man_88]both ^1+1 and ^2+1 seem to have the same digital root.
[/QUOTE]

Digital root = Number mod 9.

8 mod 9 = 1.
50 mod 9 = 5.

8=/=5.

Conjecture disproven. The law of small numbers prevails yet again.

Odd counterexample:

30+1 mod 9 = 4.
30^2+1 mod 9 = 1
4 =/= 1

[QUOTE=science_man_88;225196]both ^1+1 and ^2+1 seem to have the same digital root.[/QUOTE]

No, not usually. Run the code I posted to look at small examples -- the first half-dozen all fail.

[QUOTE=3.14159;225198]Digital root = Number mod 9.

8 mod 9 = 8.
8 * 8 mod 9 = 64 mod 9 = 1.

Conjecture disproven. The law of small numbers prevails yet again[/QUOTE]

I think he meant only the numbers that fulfill your conditions rather than just any number. But it's wrong either way.

# m is the amount of Mersenne numbers you multiply together %9 means mod 9 and the subscript is the index in the sequence for that #M .

[QUOTE=3.14159;225198]Digital root = Number mod 9.

8 mod 9 = 1.
50 mod 9 = 5.

8=/=5.

Conjecture disproven. The law of small numbers prevails yet again.

Odd counterexample:

31 mod 9 = 4.
901 mod 9 = 1
4 =/= 1[/QUOTE]

I summed the digits Pi and I got both having a additive root of 1.

[QUOTE=science_man_88]I summed the digits Pi and I got both having a additive root of 1.
[/QUOTE]

Tell me: How did you reduce infinity to 1?

[QUOTE=science_man_88;225200]# m is the amount of Mersenne numbers you multiply together %9 means mod 9 and the subscript is the index in the sequence for that #M .[/QUOTE]

Means nothing to me.

You have three columns for %9 but they have different values. I don't know why and you haven't said.

Are you really multiplying 5 different Mersenne numbers together? That's what you say, but I'd be more inclined to believe 5 as the exponent of the first of two Mersenne numbers you multiply.

And what is the x and what would allow us to find it?

[QUOTE=3.14159;225202]Tell me: How do you know the additive root of infinity?[/QUOTE]

I'm talking [TEX]\sum[/TEX](digits of both) =1

[QUOTE=science_man_88]I'm talking (digits of both) =1
[/QUOTE]

The sum of the digits of pi = 1? So 3.141592653589793238462643383279... = 1? Give us your math proof if this.

For the lulz:

[code]3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420198938095257201065485863278865936153381827968230301952035301852968995773622599413891249721775283479131515574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012858361603563707660104710181942955596198946767837449448255379774726847104047534646208046684259069491293313677028989152104752162056966024058038150193511253382430035587640247496473263914199272604269922796782354781636009341721641219924586315030286182974555706749838505494588586926995690927210797509302955321165344987202755960236480665499119881834797753566369807426542527862551818417574672890977772793800081647060016145249192173217214772350141441973568548161361157352552133475741849468438523323907394143334547762416862518983569485562099219222184272550254256887671790494601653466804988627232791786085784383827967976681454100953883786360950680064225125205117392984896084128488626945604241965285022210661186306744278622039194945047123713786960956364371917287467764657573962413890865832645995813390478027590099465764078951269468398352595709825822620522489407726719478268482601476990902640136394437455305068203496252451749399651431429809190659250937221696461515709858387410597885959772975498930161753928468138268683868942774155991855925245953959431049972524680845987273644695848653836736222626099124608051243884390451244136549762780797715691435997700129616089441694868555848406353422072225828488648158456028506016842739452267467678895252138522549954666727823986456596116354886230577456498035593634568174324112515076069479451096596094025228879710893145669136867228748940560101503308617928680920874760917824938589009714909675985261365549781893129784821682998948722658804857564014270477555132379641451523746234364542858444795265867821051141354735739523113427166102135969536231442952484937187110145765403590279934[/code]

The first 2800 digits of pi. Remember, the sum of those digits is 1! LOL!

You do realize that the only number whose sum of digits is 1 is 1, correct?

look if you multiply 3 consecutive Mersenne primes (MP[SUB]x[/SUB] * MP[SUB]x+1[/SUB] * MP[SUB]x+2[/SUB]) we get a additive root so far for primes it seems this sequence always starts with 3 if we can prove something like always starting with 3,1,4 for primes then we can use this knowledge to predict modulo of the next or further Mersenne primes given the modulo it should be able to make a smaller list of checks to do for a given range until we find the next one then we move on to search the next modulo for a given range as long as we can predict which exponents give a certain modulo we get a small list and then we can limit these to prime exponents to check.

[QUOTE=3.14159;225205]The sum of the digits of pi = 1? So 3.141592653589793238462643383279... = 1? Give us your math proof if this.

For the lulz:

[code]3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420198938095257201065485863278865936153381827968230301952035301852968995773622599413891249721775283479131515574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012858361603563707660104710181942955596198946767837449448255379774726847104047534646208046684259069491293313677028989152104752162056966024058038150193511253382430035587640247496473263914199272604269922796782354781636009341721641219924586315030286182974555706749838505494588586926995690927210797509302955321165344987202755960236480665499119881834797753566369807426542527862551818417574672890977772793800081647060016145249192173217214772350141441973568548161361157352552133475741849468438523323907394143334547762416862518983569485562099219222184272550254256887671790494601653466804988627232791786085784383827967976681454100953883786360950680064225125205117392984896084128488626945604241965285022210661186306744278622039194945047123713786960956364371917287467764657573962413890865832645995813390478027590099465764078951269468398352595709825822620522489407726719478268482601476990902640136394437455305068203496252451749399651431429809190659250937221696461515709858387410597885959772975498930161753928468138268683868942774155991855925245953959431049972524680845987273644695848653836736222626099124608051243884390451244136549762780797715691435997700129616089441694868555848406353422072225828488648158456028506016842739452267467678895252138522549954666727823986456596116354886230577456498035593634568174324112515076069479451096596094025228879710893145669136867228748940560101503308617928680920874760917824938589009714909675985261365549781893129784821682998948722658804857564014270477555132379641451523746234364542858444795265867821051141354735739523113427166102135969536231442952484937187110145765403590279934[/code]

The first 2800 digits of pi. Remember, the sum of those digits is 1! LOL!

You do realize that the only number whose sum of digits is 1 is 1, correct?[/QUOTE]

I mean summation until you get a single digit number they both equal 1.

[QUOTE=science_man_88]I mean summation until you get a single digit number they both equal 1.
[/QUOTE]

Tell me, how can you reduce infinity to 1?

Sum of digits(infinity) = infinity.

How did you avoid the circular loop?

This should be amusing :missingteeth:

[QUOTE=science_man_88;225206]look if you multiply 3 consecutive Mersenne primes (MP[SUB]x[/SUB] * MP[SUB]x+1[/SUB] * MP[SUB]x+2[/SUB]) we get a additive root so far for primes it seems this sequence always starts with 3 if we can prove something like always starting with 3,1,4 for primes then we can use this knowledge to predict modulo of the next or further Mersenne primes given the modulo it should be able to make a smaller list of checks to do for a given range until we find the next one then we move on to search the next modulo for a given range as long as we can predict which exponents give a certain modulo we get a small list and then we can limit these to prime exponents to check.[/QUOTE]

Digital roots of the product of three consecutive Mersenne primes: 3, 1, 4, 4, 4, 4, 1, 4, 7, 7, 7, 4, 4, 1, 1, 1, 4, 4, 7, 7, 1, 1, 1, 7, 4, 4, 4, 4, 1, 4, 7, 7, 7, 7, 1, 1, 7

But what do you mean, "always starts with 3"?

[QUOTE=science_man_88;225207]I mean summation until you get a single digit number they both equal 1.[/QUOTE]

I can't think of any reasonable way to define this. How are you calculating it?

[QUOTE=science_man_88;225194]my idea is this:

[CODE]#M %9[SUB]1[/SUB] %9[SUB]2[/SUB] %9[SUB]3[/SUB]
2 3 1 4 x
3 3 1 4
5 3 [/CODE]

care to use the data in the table to calculate x ?[/QUOTE]

look %9[SUB]3[/SUB] in the 3 row is 4 this is for Mp3*mp4*mp5

if we already know there's a pattern and the first 2 we can predict it for Mp5 and hence Mp4*mp5 which would give us x hence we can use the pattern (if there is one) to predict modulo even at the basic level of a single Mersenne and hence we only have to check possible Mersenne exponents that give a number that can give a number of that modulo.

@ CRG: digit one + digit 2 ....... until the end of both continue until you get one digit that digit is 1 for both in the example you gave.

Ah, I think I have it. Together with my above code take
[code]ProdK(k)=k--;vector(#A43-k,i,prod(j=0,k,drMersenne(A43[i+j]))%9)[/code]
which gives the digital roots of the product of k consecutive Mersenne numbers. So ProdK(1) gives [url=http://oeis.org/classic/A135928]A135928[/url](n), ProdK(2) gives [url=http://oeis.org/classic/A010888]A010888[/url]([url=http://oeis.org/classic/A165223]A165223[/url](n)), etc.

Then you have
[code]>for(k=1,15,print(k"\t"ProdK(k)))
1 [3, 7, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 4, 4, 4, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 4, 4, 4, 1]
2 [3, 1, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 4, 1, 1, 1, 1, 4, 4, 4, 7, 7, 7, 7, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 7, 7, 7, 4]
3 [3, 1, 4, 4, 4, 4, 1, 4, 7, 7, 7, 4, 4, 1, 1, 1, 4, 4, 7, 7, 1, 1, 1, 7, 4, 4, 4, 4, 1, 4, 7, 7, 7, 7, 1, 1, 7]
4 [3, 1, 7, 4, 4, 4, 4, 7, 7, 1, 7, 4, 4, 1, 1, 4, 4, 7, 1, 1, 4, 4, 1, 7, 7, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 1]
5 [3, 4, 7, 4, 4, 7, 7, 7, 1, 1, 7, 4, 4, 1, 4, 4, 7, 1, 4, 4, 7, 4, 1, 1, 7, 4, 4, 7, 7, 7, 1, 4, 4, 4, 4]
6 [3, 4, 7, 4, 7, 1, 7, 1, 1, 1, 7, 4, 4, 4, 4, 7, 1, 4, 7, 7, 7, 4, 4, 1, 7, 4, 7, 1, 7, 1, 4, 7, 7, 4]
7 [3, 4, 7, 7, 1, 1, 1, 1, 1, 1, 7, 4, 7, 4, 7, 1, 4, 7, 1, 7, 7, 7, 4, 1, 7, 7, 1, 1, 1, 4, 7, 1, 7]
8 [3, 4, 1, 1, 1, 4, 1, 1, 1, 1, 7, 7, 7, 7, 1, 4, 7, 1, 1, 7, 1, 7, 4, 1, 1, 1, 1, 4, 4, 7, 1, 1]
9 [3, 7, 4, 1, 4, 4, 1, 1, 1, 1, 1, 7, 1, 1, 4, 7, 1, 1, 1, 1, 1, 7, 4, 4, 4, 1, 4, 7, 7, 1, 1]
10 [3, 1, 4, 4, 4, 4, 1, 1, 1, 4, 1, 1, 4, 4, 7, 1, 1, 1, 4, 1, 1, 7, 7, 7, 4, 4, 7, 1, 1, 1]
11 [3, 1, 7, 4, 4, 4, 1, 1, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 4, 1, 1, 1, 1, 7, 7, 7, 1, 4, 1]
12 [3, 4, 7, 4, 4, 4, 1, 4, 4, 7, 7, 7, 1, 1, 1, 1, 4, 4, 4, 1, 4, 4, 1, 1, 1, 1, 4, 4]
13 [3, 4, 7, 4, 4, 4, 4, 4, 7, 1, 1, 1, 4, 1, 1, 4, 4, 4, 4, 4, 7, 4, 4, 4, 4, 4, 4]
14 [3, 4, 7, 4, 4, 7, 4, 7, 1, 4, 4, 4, 4, 1, 4, 4, 4, 4, 7, 7, 7, 7, 7, 7, 7, 4]
15 [3, 4, 7, 4, 7, 7, 7, 1, 4, 7, 7, 4, 4, 4, 4, 4, 4, 7, 1, 7, 1, 1, 1, 1, 7][/code]

[QUOTE=CRGreathouse]I can't think of any reasonable way to define this. How are you calculating it?
[/QUOTE]


See here:

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[QUOTE=CRGreathouse;225212]Ah, I think I have it. Together with my above code take
[code]ProdK(k)=k--;vector(#A43-k,i,prod(j=0,k,drMersenne(A43[i+j]))%9)[/code]
which gives the digital roots of the product of k consecutive Mersenne numbers. So ProdK(1) gives [url=http://oeis.org/classic/A135928]A135928[/url](n), ProdK(2) gives [url=http://oeis.org/classic/A010888]A010888[/url]([url=http://oeis.org/classic/A165223]A165223[/url](n)), etc.

Then you have[/QUOTE]
well wanted to keep it more specific to Mersenne primes all I can say from this is that the line of 3's going down map on diagonals to the top one.

[QUOTE=science_man_88;225215]well wanted to keep it more specific to Mersenne primes[/QUOTE]

Yes, these are (the digital roots of products of) Mersenne primes.

[QUOTE=science_man_88;225215]all I can say from this is that the line of 3's going down map on diagonals to the top one.[/QUOTE]

I trust you mean that the first column (the first number in each row) is always 3?

Can you prove it?

I think the proof is that the first 2 numbers seem to always give 3:

that means if we only deal with 3 [1,4,7]

once all those combo's appear and give 3 in the next one by my logic all that only use 3,[1,4,7] to start will always give a 3 in the next one hence is there a proof about 1,4,7 being the only ones used in most of these if so we can prove it.

So what are 3*1, 3*4, and 3*7 mod 9? You're almost there.

[QUOTE=CRGreathouse;225220]So what are 3*1, 3*4, and 3*7 mod 9? You're almost there.[/QUOTE]

they all come back to 3 so assuming my idea is correct the first column must always be 3 hence we can map it through the diagonals and get the modulo for the Mersenne primes to come.

though why multiply them in the next in the row they cover almost all the same range.

the hard part is how to map it accurately without knowing it.

[QUOTE=science_man_88;225221]they all come back to 3 so assuming my idea is correct the first column must always be 3[/QUOTE]

Right. This is a [i]proof[/i] that the first column is always 3 where it exists. (We don't know if there are infinitely many rows because we don't know if there are infinitely many Mersenne primes.)

[QUOTE=science_man_88;225221]hence we can map it through the diagonals and get the modulo for the Mersenne primes to come.[/QUOTE]

You can't tell anything about future Mersenne primes because no matter what they are, you'll get a 3. Right?

[QUOTE=CRGreathouse;225212]Ah, I think I have it. Together with my above code take
[code]ProdK(k)=k--;vector(#A43-k,i,prod(j=0,k,drMersenne(A43[i+j]))%9)[/code]
which gives the digital roots of the product of k consecutive [B][U][I][COLOR="Red"]Mersenne numbers[/COLOR][/I][/U][/B]. So ProdK(1) gives [url=http://oeis.org/classic/A135928]A135928[/url](n), ProdK(2) gives [url=http://oeis.org/classic/A010888]A010888[/url]([url=http://oeis.org/classic/A165223]A165223[/url](n)), etc.

Then you have[/quote]

you say Mersenne numbers ... any other patterns in it if it is for Mersenne primes such that it would help predict modulo ?

can this be turned into a triangle by turning it 45 degrees ?

I see more weird things in the diagonals that would become rows of this triangle not saying they tell much lol.

[QUOTE=science_man_88;225224]you say Mersenne numbers ... any other patterns in it if it is for Mersenne primes such that it would help predict modulo ?[/QUOTE]

I'm sorry, I meant Mersenne primes. The same pattern would hold for Mersenne numbers with exponents relatively prime to 6.

You *know* that you can't predict anything about any sequence that has exponents that are relatively prime to 6, because you're already shown how those sequences have to start. It cuts both ways: because the proof shows that all such sequences will have the same behavior, you can't use that behavior to predict anything within those sequences.

[QUOTE=science_man_88;225228]can this be turned into a triangle by turning it 45 degrees ?[/QUOTE]

Sure, whatever you want. Feel free to modify my code as needed.

[code]>for(k=1,15,print(k"\t"ProdK(k)))
1 [3, 7, [COLOR="Red"]4[/COLOR], 1, 1, [COLOR="Yellow"]4[/COLOR], 1, 1, 1, 4, 4, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 4, 4, 4, 4, 1, 1, 4, 1, 1, 1, 4, 4, 1, 4, 4, 4, 4, 1]
2 [3, [COLOR="red"]1, 4,[/COLOR] 1, [COLOR="yellow"]4, 4,[/COLOR] 1, 1, 4, 7, 4, 4, 4, 1, 1, 1, 1, 4, 4, 4, 7, 7, 7, 7, 4, 1, 4, 4, 1, 1, 4, 7, 4, 4, 7, 7, 7, 4]
3 [[COLOR="red"]3, 1, 4,[/COLOR][COLOR="yellow"] 4, 4, 4, [/COLOR]1, 4, 7, 7, 7, 4, 4, 1, 1, 1, 4, 4, 7, 7, 1, 1, 1, 7, 4, 4, 4, 4, 1, 4, 7, 7, 7, 7, 1, 1, 7]
4 [[COLOR="Red"]3, 1,[/COLOR] [COLOR="Yellow"]7, 4, 4, 4[/COLOR], 4, 7, 7, 1, 7, 4, 4, 1, 1, 4, 4, 7, 1, 1, 4, 4, 1, 7, 7, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 1]
5 [[COLOR="Red"]3[/COLOR], [COLOR="yellow"]4, 7, 4, 4[/COLOR], 7, 7, 7, 1, 1, 7, 4, 4, 1, 4, 4, 7, 1, 4, 4, 7, 4, 1, 1, 7, 4, 4, 7, 7, 7, 1, 4, 4, 4, 4]
6 [[COLOR="Yellow"]3, 4, 7, 4[/COLOR], 7, 1, 7, 1, 1, 1, 7, 4, 4, 4, 4, 7, 1, 4, 7, 7, 7, 4, 4, 1, 7, 4, 7, 1, 7, 1, 4, 7, 7, 4]
7 [[COLOR="Yellow"]3, 4, 7[/COLOR], 7, 1, 1, 1, 1, 1, 1, 7, 4, 7, 4, 7, 1, 4, 7, 1, 7, 7, 7, 4, 1, 7, 7, 1, 1, 1, 4, 7, 1, 7]
8 [[COLOR="yellow"]3, 4[/COLOR], 1, 1, 1, 4, 1, 1, 1, 1, 7, 7, 7, 7, 1, 4, 7, 1, 1, 7, 1, 7, 4, 1, 1, 1, 1, 4, 4, 7, 1, 1]
9 [[COLOR="Yellow"]3[/COLOR], 7, 4, 1, 4, 4, 1, 1, 1, 1, 1, 7, 1, 1, 4, 7, 1, 1, 1, 1, 1, 7, 4, 4, 4, 1, 4, 7, 7, 1, 1]
10 [3, 1, 4, 4, 4, 4, 1, 1, 1, 4, 1, 1, 4, 4, 7, 1, 1, 1, 4, 1, 1, 7, 7, 7, 4, 4, 7, 1, 1, 1]
11 [3, 1, 7, 4, 4, 4, 1, 1, 4, 4, 4, 4, 7, 7, 1, 1, 1, 4, 4, 1, 1, 1, 1, 7, 7, 7, 1, 4, 1]
12 [3, 4, 7, 4, 4, 4, 1, 4, 4, 7, 7, 7, 1, 1, 1, 1, 4, 4, 4, 1, 4, 4, 1, 1, 1, 1, 4, 4]
13 [3, 4, 7, 4, 4, 4, 4, 4, 7, 1, 1, 1, 4, 1, 1, 4, 4, 4, 4, 4, 7, 4, 4, 4, 4, 4, 4]
14 [3, 4, 7, 4, 4, 7, 4, 7, 1, 4, 4, 4, 4, 1, 4, 4, 4, 4, 7, 7, 7, 7, 7, 7, 7, 4]
15 [3, 4, 7, 4, 7, 7, 7, 1, 4, 7, 7, 4, 4, 4, 4, 4, 4, 7, 1, 7, 1, 1, 1, 1, 7][/code]

here's one such thing I see.

SLSN, for that one. Since you're going along the antidiagonals there aren't any more terms to check...

[QUOTE=CRGreathouse;225234]SLSN, for that one. Since you're going along the antidiagonals there aren't any more terms to check...[/QUOTE]

SLSN ?

Richard Guy's Strong Law of Small Numbers.

[QUOTE=CRGreathouse]Richard Guy's Strong Law of Small Numbers.
[/QUOTE]

He bases everything on the law of small numbers. I might as well begin calling him "The law of small numbers guy."

[QUOTE=3.14159;225237]He bases everything on the law of small numbers. I might as well begin calling him "The law of small numbers guy."[/QUOTE]

I base every thing I post on things I know they just happen to run into it more often than [U]not[/U] lol.

underlined = haven't found one yet.