OFFSET
1,2
COMMENTS
From Antti Karttunen, Jul 31 - Aug 06 2020: (Start)
Question: is it possible that a(k) = 2*k for any k? If not, then the deficiency (A033879) cannot be -1, and there are no quasiperfect numbers. If there were such cases, then A156552(k) = q would be an instance of quasiperfect number, which should also be an odd square, thus k would need to be of the form 4u+2.
In range n <= 10000, a(n) is a nontrivial multiple of n only at n = [25, 35, 343, 539, 847, 3315] with a(n) = [125, 105, 2401, 2695, 2541, 9945]. The quotients are thus also odd: 5, 3, 7, 5, 3, 3.
This rather meager empirical evidence motivates a conjecture that no quotient a(n)/n may be an even integer, and particularly, never a power of 2 larger than one, which (when translated back to the ordinary, unconjugated sigma) claims that it is not possible that sigma(n) = 2^k * n + 2^k - 1, for any n > 1, k > 0. See also A336700 and A336701, where this leads to a rather surprising empirical observation.
(End)
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000 (computed using Hans Havermann's factorization of A156552)
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
FORMULA
PROG
(PARI)
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
(PARI) A332223(n) = if(1==n, n, A005940(1+sumdiv(A156552(n), d, d))); \\ Antti Karttunen, Aug 04 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 12 2020
STATUS
approved