A046840 -id:A046840 - cp's OEIS frontend

A046839 Numbers k such that the number of divisors of k divides the sum of cubes of divisors of k.

1, 3, 5, 6, 7, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 27, 29, 30, 31, 33, 35, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 51, 53, 54, 55, 56, 57, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 73, 77, 78, 79, 83, 85, 86, 87, 89, 91, 92, 93, 94, 95, 96, 97, 99, 101, 102, 103

Offset: 1

Labos Elemer

nonn

The first 42 terms agree with A003601 but a(43) = 64 is not a term in A003601.

			64 is a term since it has 7 divisors, and sigma_3(64) = 299593 = 7 * 42799 is divisble by 7.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)

Cf. A000005, A001158.

Cf. A003601, A020486, A046840.

[n: n in [1..110] | IsZero(DivisorSigma(3, n) mod NumberOfDivisors(n))]; // Bruno Berselli, Apr 11 2013

Select[Range[103], Divisible[DivisorSigma[3, #], DivisorSigma[0, #]] &] (* Jayanta Basu, Jun 29 2013 *)

isok(n) = sigma(n, 3) % numdiv(n) == 0; \\ Michel Marcus, May 13 2018

A361683 a(n) is the least k such that tau(k) divides sigma_n(k) but not sigma(k), or -1 if no such k exists.

Original entry on oeis.org

4, 64, 4, 7168, 4, 606528, 4, 64, 4, 4194304, 4

Offset: 2

Mohammed Yaseen, Mar 20 2023

a(13) <= 31525197391593472. - David A. Corneth, Mar 20 2023
From Thomas Scheuerle, Mar 22 2023: (Start)
a(17) <= 15211807202738752817960438464512 and a(19) <= 2^190*11.
Conjecture: a(n) is of the form 2^b*p1^c*p2^d*...*pk^j with b > 0 and A020639(n) divides b*(c+1)*(d+1)*...*(j+1). (p1, p2, ..., pk are distinct odd prime numbers). (End)

Cf. A003601, A020486, A046839, A046840.
Cf. A000005, A000203, A001157, A001158.
Cf. A020639, A010709 (bisection).

a[n_] := Module[{k = 1, d}, While[Divisible[DivisorSigma[1, k], (d = DivisorSigma[0, k])] || !Divisible[DivisorSigma[n, k], d], k++]; k]; Array[a, 11, 2] (* Amiram Eldar, Mar 20 2023 *)

isok(k, n) = my(f=factor(k), nd=numdiv(f)); (sigma(f) % nd) && !(sigma(f,n) % nd);
a(n) = my(k=1); while (!isok(k,n), k++); k; \\ Michel Marcus, Mar 20 2023

a(2*m) = 4 for m >= 1.
a(6*m-3) = 64 for m >= 1.
From Thomas Scheuerle, Mar 22 2023: (Start)
a(m) <= a(A020639(m)) if a(A020639(m)) <> -1.
Conjecture: For primes q > p, a(q) > a(p). If true, we could replace "<=" with "=" in the above formula. (End)

cp's OEIS Frontend

A046839 Numbers k such that the number of divisors of k divides the sum of cubes of divisors of k.

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