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URL: https://oeis.org/A065960

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A065960
a(n) = n^4*Product_{distinct primes p dividing n} (1+1/p^4).
15
1, 17, 82, 272, 626, 1394, 2402, 4352, 6642, 10642, 14642, 22304, 28562, 40834, 51332, 69632, 83522, 112914, 130322, 170272, 196964, 248914, 279842, 356864, 391250, 485554, 538002, 653344, 707282, 872644, 923522, 1114112, 1200644
OFFSET
1,2
LINKS
F. A. Lewis and others, Problem 4002, Amer. Math. Monthly, Vol. 49, No. 9, Nov. 1942, pp. 618-619.
FORMULA
Multiplicative with a(p^e) = p^(4*e)+p^(4*e-4). - Vladeta Jovovic, Dec 09 2001
a(n) = n^4 * Sum_{d|n} mu(d)^2/d^4. - Benoit Cloitre, Apr 07 2002
a(n) = J_8(n)/J_4(n) = A069093(n)/A059377(n), where J_k is the k-th Jordan Totient Function. - Enrique Pérez Herrero, Aug 29 2010
Dirichlet g.f.: zeta(s)*zeta(s-4)/zeta(2*s). - R. J. Mathar, Jun 06 2011
From Vaclav Kotesovec, Sep 19 2020: (Start)
Sum_{k=1..n} a(k) ~ 18711*zeta(5)*n^5 / Pi^10.
Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^4/(p^8-1)) = 1.078178802583045599985995264729541574821218371712364313741065126120993131... (End)
MAPLE
A065960 := proc(n) n^4*mul(1+1/p^4, p=numtheory[factorset](n)) ; end proc:
seq(A065960(n), n=1..20) ; # R. J. Mathar, Jun 06 2011
MATHEMATICA
a[n_] := n^4*DivisorSum[n, MoebiusMu[#]^2/#^4&]; Array[a, 40] (* Jean-François Alcover, Dec 01 2015 *)
f[p_, e_] := p^(4*e) + p^(4*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 12 2020 *)
PROG
(PARI) for(n=1, 100, print1(n^4*sumdiv(n, d, moebius(d)^2/d^4), ", "))
CROSSREFS
Sequences of the form n^k * Product_ {p|n, p prime} (1 + 1/p^k) for k=0..10: A034444 (k=0), A001615 (k=1), A065958 (k=2), A065959 (k=3), this sequence (k=4), A351300 (k=5), A351301 (k=6), A351302 (k=7), A351303 (k=8), A351304 (k=9), A351305 (k=10).
Sequence in context: A088687 A321560 A034678 * A017671 A001159 A053820
KEYWORD
nonn,mult,easy
AUTHOR
N. J. A. Sloane, Dec 08 2001
STATUS
approved