A336700 Numbers k such that the odd part of (1+k) divides (1 + odd part of sigma(k)).

1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 2431, 2943, 3775, 4095, 8191, 13311, 14335, 16383, 17407, 21951, 22527, 32767, 34335, 44031, 57855, 65535, 85375, 131071, 204799, 262143, 376831, 524287, 923647, 1048575, 1562623, 1632255, 2056191, 2097151, 2744319, 4194303, 6815743, 8388607, 8781823, 10059775, 16777215

Offset: 1

Author

Antti Karttunen, Aug 02 2020

Keywords

• nonn

Comments

Numbers k for which A337194(k) = 1+A161942(k) is a multiple of A000265(1+k).

Conjecture: After 1, all terms are of the form 4u+3 (in A004767). If this could be proved, then it would refute at once the existence of both the odd perfect numbers and the quasiperfect numbers. Concentrating on the latter is probably easier, as it is known that all quasiperfect numbers must be odd squares, thus k is of the form 4u+1, in which case the condition given in A336701 that A000265(1+A000265(sigma(k))) must be equal to A000265(1+k) reduces to a simpler form, A000265(1+sigma(k)) = (1+k)/2, and as k = s^2, with s odd, so (s^2 + 1)/2 should divide 1+sigma(s^2). Does that condition allow any other solutions than s=1 ? See A337339.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..77; all terms < 2^32
• Paolo Cattaneo, Sui numeri quasiperfetti, Bollettino dell’Unione Matematica Italiana, Serie 3, Vol.6(1951), n.1, p. 59-62.
• P. Hagis and G. L. Cohen, Some Results Concerning Quasiperfect Numbers, J. Austral. Math. Soc. Ser. A 33, 275-286, 1982.
• V. Siva Rama Prasad and C. Sunitha, On quasiperfect numbers, Notes on Number Theory and Discrete Mathematics, Vol. 23, 2017, No. 3, 73-78.
• Eric Weisstein's World of Mathematics, Quasiperfect Number
• Index entries for sequences where odd perfect numbers must occur, if they exist at all
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A000203, A000265, A004767, A016754, A161942, A228058, A336697, A336698, A337339, A337342.

Subsequences: A000225, A336701 (terms where the quotient is a power of 2).

Programs

Mathematica

Block[{f}, f[n_] := n/2^IntegerExponent[n, 2]; Select[Range[2^20], Mod[f[1 + f[DivisorSigma[1, #]]], f[1 + #]] == 0 &] ] (* Michael De Vlieger, Aug 22 2020 *)

PARI

A000265(n) = (n>>valuation(n,2));
isA336700(n) = !((1+A000265(sigma(n)))%A000265(1+n));