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a(n) = t(n,3) = Sum_{d|n} tau(d^3), where tau(n) = number of divisors of n, cf. A000005.
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%I #47 Aug 20 2021 16:57:03

%S 1,5,5,12,5,25,5,22,12,25,5,60,5,25,25,35,5,60,5,60,25,25,5,110,12,25,

%T 22,60,5,125,5,51,25,25,25,144,5,25,25,110,5,125,5,60,60,25,5,175,12,

%U 60,25,60,5,110,25,110,25,25,5,300,5,25,60,70,25,125,5,60,25,125,5,264

%N a(n) = t(n,3) = Sum_{d|n} tau(d^3), where tau(n) = number of divisors of n, cf. A000005.

%C Inverse Mobius transform of A048785. - _R. J. Mathar_, Feb 09 2011

%H Antti Karttunen, <a href="/A061391/b061391.txt">Table of n, a(n) for n = 1..10000</a>

%F t(n, k) = Sum_{d|n} tau(d^k) is multiplicative: if the canonical factorization of n = Product p^e(p) over primes then t(n, k) = Product t(p^e(p), k), t(p^e(p), k) = (1/2) *(k*e(p)+2)*(e(p)+1).

%F For k=2 we get an interesting identity: Sum_{d|n} tau(d^2)=(tau(n))^2, cf. A048691, A035116.

%F a(n) = Sum_{d|n} tau(n*d). - _Benoit Cloitre_, Nov 30 2002

%F G.f.: Sum_{n>=1} tau(n^3)*x^n/(1-x^n). - _Joerg Arndt_, Jan 01 2011

%F Dirichlet g.f.: zeta(s)^3 * Product_{primes p} (1 + 2/p^s). - _Vaclav Kotesovec_, May 15 2021

%F Dirichlet g.f.: zeta(s)^5 * Product_{primes p} (1 - 3/p^(2*s) + 2/p^(3*s)). - _Vaclav Kotesovec_, Aug 20 2021

%t f[p_, e_] := (3*e^2 + 5*e + 2)/2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 16 2020 *)

%o (PARI)

%o A061391 = n -> sumdiv(n, d, numdiv(d^3));

%o for(n=1, 10000, write("b061391.txt", n, " ", A061391(n)));

%o \\ _Antti Karttunen_, Jan 17 2017

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + 2*X)/(1 - X)^3)[n], ", ")) \\ _Vaclav Kotesovec_, May 15 2021

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - 3*X^2 + 2*X^3)/(1 - X)^5)[n], ", ")) \\ _Vaclav Kotesovec_, Aug 20 2021

%Y Cf. t(n, 0) = A000005(n), t(n, 1) = A007425(n), t(n, 2) = A035116(n).

%Y Cf. A048691.

%K nonn,mult

%O 1,2

%A _Vladeta Jovovic_, Apr 29 2001