A332223 a(1) = 1, and for n > 1, a(n) = A005940(1+sigma(A156552(n))).

1, 2, 4, 5, 8, 9, 16, 7, 25, 18, 32, 25, 64, 21, 21, 49, 128, 27, 256, 35, 40, 121, 512, 49, 125, 385, 49, 121, 1024, 13, 2048, 13, 225, 1573, 105, 77, 4096, 57, 187, 343, 8192, 63, 16384, 65, 55, 4693, 32768, 121, 625, 32, 15625, 85, 65536, 81, 180, 91, 253, 9945, 131072, 175, 262144, 508079, 625, 847, 729, 169, 524288, 2057, 2601, 105, 1048576

Offset: 1

Author

Antti Karttunen, Feb 12 2020

Keywords

• nonn

Comments

From Antti Karttunen, Jul 31 - Aug 06 2020: (Start)

As a curiosity, like with sigma, also here a(14) = a(15). [Cf. also A003973 and A341512]

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
• Index entries for sequences computed from indices in prime factorization
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A000203, A005940, A027687, A156552, A246277, A323243, A324201, A324899, A324909, A332224, A332225, A332463 (Möbius transform).

Cf. A003961, A332449, A332450, A332451, A332460 (for other functions similarly conjugated).

Cf. also A003973, A033879, A326042, A332222, A336700, A336701, A341512.

Programs

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
A332223(n) = if(1==n,n,A005940(1+sigma(A156552(n))));

PARI

A332223(n) = if(1==n,n,A005940(1+sumdiv(A156552(n),d,d))); \\ Antti Karttunen, Aug 04 2020

Formula

For n > 1, a(n) = A005940(1+A000203(A156552(n))) = A005940(1+A323243(n)).

a(A324201(n)) = A003961(A324201(n)). [It's an open problem whether A324201 gives all such solutions]

For n > 1, a(n) = A005940(1 + (Sum_{d|A156552(n)} d)). - Antti Karttunen, Aug 04 2020