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A332223
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a(1) = 1, and for n > 1, a(n) = A005940(1+sigma(A156552(n))).
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10
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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
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OFFSET
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1,2
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COMMENTS
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As a curiosity, like with sigma, also here a(14) = a(15).
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)
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LINKS
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V. Siva Rama Prasad and C. Sunitha, On quasiperfect numbers, Notes on Number Theory and Discrete Mathematics, Vol. 23, 2017, No. 3, 73-78.
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FORMULA
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PROG
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(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
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CROSSREFS
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Cf. A000203, A005940, A027687, A156552, A246277, A323243, A324201, A324899, A324909, A332224, A332225, A332463 (Möbius transform).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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