OFFSET
1,2
COMMENTS
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
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 *)
PROG
KEYWORD
nonn,more
AUTHOR
Antti Karttunen, Aug 02 2020
STATUS
approved