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.

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, 58, 59, 63, 64, 68, 74, 75, 77, 82, 83, 92, 93, 94, 110, 116, 119, 122, 125, 128, 129, 130, 131, 134, 136, 137, 142, 145, 152, 161, 164, 170, 171, 184, 185, 189, 194

Offset: 1

Author

Simon Wundling, Mar 13 2022

Keywords

• nonn

Comments

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.

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.

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.

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.

Examples

			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.
		

Crossrefs

Subsequence of A005100.

Cf. A000203, A033879, A005101, A000396.

Cf. A000079, A237588, A019434.

Programs

PARI

f(n) = 2*n - sigma(n);
isok(m) = while (1, m = f(m); if (m==1, return(1)); if (m<=0, return(0));); \\ Michel Marcus, Mar 13 2022