0,1

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?

Toby Bailey and Chris Smyth Primitive solutions of n|2^n+1.(This is the same link as at A136473.)

The Cunningham Project

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.

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 5-th '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.

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);

Cf. A006521, A136473, A136474.

Sequence in context: A095106 A130184 A096935 this_sequence A074249 A101617 A131443

Adjacent sequences: A136472 A136473 A136474 this_sequence A136476 A136477 A136478

nonn,tabf

Chris Smyth (c.smyth(AT)ed.ac.uk), Feb 16 2008, Feb 19 2008, Feb 22 2008