cp's OEIS Frontend

A358239 Numbers k such that the aliquot sequence of 2^k ends with the prime 3.

%I A358239 #29 Nov 13 2022 04:11:49
%S A358239 2,4,55,164,305,317
%N A358239 Numbers k such that the aliquot sequence of 2^k ends with the prime 3.
%H A358239 Jean-Luc Garambois, <a href="http://www.aliquotes.com/aliquotes_puissances_entieres/aliquotes_puissances_entieres.html">Aliquot sequences starting on integer powers n^i</a>, n^i project.
%H A358239 Mersenne forum, <a href="https://mersenneforum.org/showpost.php?p=617134&amp;postcount=2053">n^i project</a>.
%F A358239 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.
%e A358239 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.
%o A358239 (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););
%o A358239 isok(m) = f(m) == 3; \\ _Michel Marcus_, Nov 05 2022
%Y A358239 Cf. A127163, A358266.
%K A358239 nonn,hard,more
%O A358239 1,1
%A A358239 _Jean Luc Garambois_, Nov 04 2022