A336701 Numbers k for which A000265(1+A000265(sigma(k))) is equal to A000265(1+k).

1, 3, 7, 15, 31, 127, 1023, 8191, 34335, 57855, 131071, 524287, 2147483647

Offset: 1

Author

Antti Karttunen, Aug 02 2020

Keywords

• nonn
• more

Comments

Numbers k such that A336698(k) [= A000265(1+A161942(k))] is equal to A000265(1+k).

Numbers k such that A337194(k) = 2^e * A000265(1+k), for some e >= 1, where that e = A337195(k).

Any odd perfect number would trivially satisfy this condition.

Also, all hypothetical quasiperfect numbers, numbers k that satisfy sigma(k) = 2k+1, would be members.

Question: Is A066175 a subsequence of this sequence?

From Antti Karttunen, Aug 23 2020: (Start)

Numbers k such that (1+k) = 2^e * A336698(k), for some e >= 0.

Thus numbers k such that for some e >= 0, (1+k) = 2^(e-A337195(k)) * A337194(k), or equally, that A337194(k) = 2^(A337195(k)-e) * (1+k).

Conjecture: There are no even terms. This is equivalent to claim that there are no k such that A336698(k) = 1+k: If we assume that k is even, then in above equations we set e=0, and the requirement will then become that A337194(k) = 2^A337195(k)*(1+k), thus 1+k = A336698(k) = A000265(1+A000265(sigma(k))).

(End)

Links

• 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

Subsequence of A336700.

Cf. A000203, A000265, A066175, A161942, A336698, A336699, A336937, A337194, A337195.

Cf. A000668 (a subsequence).

See also comments in A326042, A332223.

Programs

Mathematica

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

PARI

A000265(n)  = (n>>valuation(n,2));
isA336701(n) = (A000265(1+A000265(sigma(n))) == A000265(1+n));