[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 ?