A358239 Numbers k such that the aliquot sequence of 2^k ends with the prime 3.
2, 4, 55, 164, 305, 317
Offset: 1
Examples
a(3)=55 because the aliquot sequence that starts with the integer 2^55 ends with the prime number 3 and there are only two smaller powers of 2 that do the same: 2^2 and 2^4.
Links
- Jean-Luc Garambois, Aliquot sequences starting on integer powers n^i, n^i project.
- Mersenne forum, n^i project.
Programs
-
PARI
f(n) = if (n==1, return(2)); my(list = List(), s=2^n); for (i=1, oo, s = sigma(s) - s; if (#select(x->(x==s), list), return(0)); if (isprime(s), return (s)); listput(list, s);); isok(m) = f(m) == 3; \\ Michel Marcus, Nov 05 2022
Formula
Define s(i) = sigma(i) - i = A000203(i) - i. Then k is a term of this sequence if the aliquot sequence obtained by repeatedly applying the mapping i->s(i) taking as initial value 2^k terminates in the prime 3.