A136475 Irregular triangle read by rows: row n gives prime factors of (2^(3^(n+1))+1)/(2^(3^n)+1).

3, 3, 19, 3, 87211, 3, 163, 135433, 272010961, 3, 1459, 139483, 10429407431911334611, 918125051602568899753, 3, 227862073, 3110690934667, 216892513252489863991753, 1102099161075964924744009, 393063301203384521164229656203691748263012766081190297429488962985651210769817

Offset: 0

Author

Christopher J. Smyth, Feb 16 2008

Keywords

• nonn
• tabf

Comments

1. The motivation for this sequence is to quickly generate integers n such that n divides 2^n+1 (sequences A006521, A136473). From the link, it is known that if 3^k||n with n|2^n+1 and n not a power of 3, then n is divisible by a prime p dividing 2^(3^k)+1. Thus for any fixed k every n with n|2^n+1 not a power of 3 is divisible by one of the following numbers: 3^k or some 3^j*p, where p>3 is a prime in A136475 before the k-th '3' and j is the number of '3's before p in the sequence.

2. Note: (2^(3^(k+1))+1)/(2^(3^k)+1) = 2^(2*3^k) - 2^(3^k) + 1.

3. For the primes dividing 2^(3^k)+1 for some k see A136474.

4. Are these numbers always squarefree?

Examples

			1. (2^(3^4)+1)/(2^(3^3)+1) = 3*163*135433*272010961, the factorization starting at the 4th '3' and ending just before the 5th '3'.
2. From Comment 1 below and k=5, we see that every n not a power of 3 satisfying n|2^n+1 (sequences A006521, A136473) is divisible by 3^5 or 3^2*19 or 3^3*87211 or 3^4*163 or 3^4*135433 or 3^4*272010961.
		

Links

• Toby Bailey and Chris Smyth Primitive solutions of n|2^n+1.(This is the same link as at A136473.)
• S. S. Wagstaff, Jr., The Cunningham Project

Crossrefs

Cf. A006521, A136473, A136474.

Programs

Maple

S:=[];for k from 0 to 4 do f:=op(2,ifactors((2^(3^(k+1))+1)/(2^(3^k)+1)));T:=[];for j to nops(f) do T:=[op(T),op(1,op(j,f))];od;S:=[op(S),op(sort(T))];od;op(S);

Formula

The prime factors of (2^(3^(k+1))+1)/(2^(3^k)+1) are given in ascending order *for each k*. For each new value of k the factorization starts with a '3', thus delimiting the different factorizations.