A358266 Numbers k such that the aliquot sequence of 2^k ends with the prime 7.
3, 10, 12, 141, 278, 387, 421
Offset: 1
Examples
a(4)=141 because the aliquot sequence that starts with the integer 2^141 ends with the prime number 7 and there are only three smaller powers of 2 that do the same: 2^3, 2^10 and 2^12.
Links
- Jean-Luc Garambois, Aliquot sequences starting on integer powers n^i, n^i project.
- Mersenne forum, n^i project.
Programs
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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) == 7;
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 7.