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a(n) = n^6 * Product_{p|n, p prime} (1 + 1/p^6).
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%I #14 Feb 12 2022 09:26:45

%S 1,65,730,4160,15626,47450,117650,266240,532170,1015690,1771562,

%T 3036800,4826810,7647250,11406980,17039360,24137570,34591050,47045882,

%U 65004160,85884500,115151530,148035890,194355200,244156250,313742650,387951930,489424000,594823322,741453700

%N a(n) = n^6 * Product_{p|n, p prime} (1 + 1/p^6).

%C Sum of the 6th powers of the divisor complements of the squarefree divisors of n.

%H Sebastian Karlsson, <a href="/A351301/b351301.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} d^6 * mu(n/d)^2.

%F a(n) = n^6 * Sum_{d|n} mu(d)^2 / d^6.

%F Multiplicative with a(p^e) = p^(6*e) + p^(6*e-6). - _Sebastian Karlsson_, Feb 08 2022

%F From _Vaclav Kotesovec_, Feb 12 2022: (Start)

%F Dirichlet g.f.: zeta(s)*zeta(s-6)/zeta(2*s).

%F Sum_{k=1..n} a(k) ~ n^7 * zeta(7) / (7 * zeta(14)) = 2606175 * n^7 * zeta(7) / (2 * Pi^14).

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^6/(p^12-1)) = 1.01709659289559607702424749979498914920118274875188346777424441790304... (End)

%t f[p_, e_] := p^(6*e) + p^(6*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* _Amiram Eldar_, Feb 08 2022 *)

%o (PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^6);

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^6*X))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 12 2022

%Y Cf. A008683 (mu).

%Y 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), A065960 (k=4), A351300 (k=5), this sequence (k=6), A351302 (k=7), A351303 (k=8), A351304 (k=9), A351305 (k=10).

%K nonn,mult

%O 1,2

%A _Wesley Ivan Hurt_, Feb 06 2022