A034677 Sum of cubes of unitary divisors of n.
1, 9, 28, 65, 126, 252, 344, 513, 730, 1134, 1332, 1820, 2198, 3096, 3528, 4097, 4914, 6570, 6860, 8190, 9632, 11988, 12168, 14364, 15626, 19782, 19684, 22360, 24390, 31752, 29792, 32769, 37296, 44226, 43344, 47450, 50654, 61740, 61544, 64638, 68922, 86688, 79508
Offset: 1
Examples
The unitary divisors of 6 are 1, 2, 3 and 6, so a(6) = 252.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
Programs
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Mathematica
scud[n_]:=Total[Select[Divisors[n],CoprimeQ[#,n/#]&]^3]; Array[scud,40] (* Harvey P. Dale, Oct 16 2016 *) f[p_, e_] := p^(3*e)+1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 14 2020 *)
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PARI
A034677_vec(len)={ a000012=direuler(p=2,len, 1/(1-X)) ; a000578=direuler(p=2,len, 1/(1-p^3*X)) ; a000578x=direuler(p=2,len, 1-p^3*X^2) ; dirmul(dirmul(a000012,a000578),a000578x) } A034677_vec(70) /* via D.g.f., R. J. Mathar, Mar 05 2011 */
Formula
Dirichlet g.f.: zeta(s)*zeta(s-3)/zeta(2s-3). - R. J. Mathar, Mar 04 2011
If n = Product (p_j^k_j) then a(n) = Product (1 + p_j^(3*k_j)). - Ilya Gutkovskiy, Nov 04 2018
Sum_{k=1..n} a(k) ~ Pi^4 * n^4 / (360 * Zeta(5)). - Vaclav Kotesovec, Feb 01 2019
Comments