PARI's commands

[CRGreathouse]

Originally Posted by 3.14159

 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)))

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?

Can you show me your code for t?

Also, you're computing t(n*m!+1) twice, which you can avoid like so:

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)
		)
	)
};

d(a,n,x,m)={
	for(n=a,x,
		if(t(n*m!+1)!=n,
			print("trivial composite","factor is",t(n*m!+1))
		,
			print(n*m!+1)
		)
	)
};

using the form

if(condition, do_if_true, do_if_false)

rather than

if(condition, do_if_true);if(!condition, do_if_false)

[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.

[3.14159]

Originally Posted by 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.

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

[CRGreathouse]

Originally Posted by 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.

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

[science_man_88]

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.

[CRGreathouse]

Originally Posted by science_man_88

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

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.

[CRGreathouse]

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.

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)

[science_man_88]

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.

[CRGreathouse]

Originally Posted by science_man_88

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.

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.

[3.14159]

@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

[science_man_88]

my idea is this:

#M  %91  %92  %93
2    3    1    4    x
3    3    1    4
5    3

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