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A352337
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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.
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0
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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