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A136475
Irregular triangle read by rows: row n gives prime factors of (2^(3^(n+1))+1)/(2^(3^n)+1).
3
3, 3, 19, 3, 87211, 3, 163, 135433, 272010961, 3, 1459, 139483, 10429407431911334611, 918125051602568899753, 3, 227862073, 3110690934667, 216892513252489863991753, 1102099161075964924744009, 393063301203384521164229656203691748263012766081190297429488962985651210769817
OFFSET
0,1
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?
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
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.
EXAMPLE
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.
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);
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved