cp's OEIS Frontend

A352337 Numbers m such that A_m(j) = 1 for some value of j, where A_m(k+1) = 2*A_m(k) - sigma(A_m(k)), sigma(m) = sum of the divisors of m, and A_m(1) = m.

%I A352337 #36 Mar 24 2022 03:41:02
%S A352337 1,2,3,4,5,8,9,10,11,14,16,17,21,22,23,26,27,32,34,35,38,39,44,55,57,
%T A352337 58,59,63,64,68,74,75,77,82,83,92,93,94,110,116,119,122,125,128,129,
%U A352337 130,131,134,136,137,142,145,152,161,164,170,171,184,185,189,194
%N A352337 Numbers m such that A_m(j) = 1 for some value of j, where A_m(k+1) = 2*A_m(k) - sigma(A_m(k)), sigma(m) = sum of the divisors of m, and A_m(1) = m.
%C A352337 If m is deficient, then 2m > sigma(m) (see A005100) and the deficiency of m is defined as 2m - sigma(m) (see A033879). Now you can check if the deficiency is also deficient and generalize this with A_m(k+1) = 2*A_m(k) - sigma(A_m(k)) and A_m(1) = m. If A_m(j) = 1 for some value of j, then m is in this sequence.
%C A352337 This sequence is a subsequence of A005100 (deficient numbers), because if m is abundant or perfect (see A005101 and A000396) then A_m(2) = 2*m - sigma(m) <= 0 instantly.
%C A352337 Since it is conjectured that 2m - sigma(m) = 1 only for m which are powers of two (see comments at A237588) all numbers in this sequence must have one k for which A_m(k) is a power of two.
%C A352337 Because of 2*2^k - sigma(2^k) = 1 all powers of two are in this sequence and with that this sequence has infinitely many terms. Further all Fermat primes (see A019434) are also in this sequence.
%e A352337 11 is in this sequence because A_11(1) = 11, A_11(2) = 2*11-(1+11) = 10, A_11(3) = 2*10-(1+2+5+10) = 2, A_11(4) = 2*2-(2+1) = 1.
%o A352337 (PARI) f(n) = 2*n - sigma(n);
%o A352337 isok(m) = while (1, m = f(m); if (m==1, return(1)); if (m<=0, return(0));); \\ _Michel Marcus_, Mar 13 2022
%Y A352337 Subsequence of A005100.
%Y A352337 Cf. A000203, A033879, A005101, A000396.
%Y A352337 Cf. A000079, A237588, A019434.
%K A352337 nonn
%O A352337 1,2
%A A352337 _Simon Wundling_, Mar 13 2022