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%I #20 Jan 25 2024 17:06:06
%S 1,10,11,47,13,110,15,158,60,130,19,517,21,150,143,441,25,600,27,611,
%T 165,190,31,1738,92,210,244,705,37,1430,39,1098,209,250,195,2820,45,
%U 270,231,2054,49,1650,51,893,780,310,55,4851,132,920,275,987,61,2440,247,2370,297,370,67,6721,69
%N a(n) = n * Sum_{d|n} tau(d)^3 / d, where tau(n) is the number of divisors of n.
%H László Tóth, <a href="http://arxiv.org/abs/1310.7053">Multiplicative arithmetic functions of several variables: a survey</a>, arXiv preprint arXiv:1310.7053 [math.NT], 2013.
%F G.f.: Sum_{k >= 1} tau(k)^3 * x^k/(1 - x^k)^2.
%F If p is prime, a(p) = 8 + p.
%F Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2)^4 * Product_{p prime} (1 + 4/p^2 + 1/p^4) = 31.237542262502... . - _Amiram Eldar_, Dec 22 2023
%F From _Peter Bala_, Jan 25 2024: (Start)
%F a(n) = Sum_{d|n, e|n} gcd(d, e) * tau(n/d) * tau(n/e) (the sum is a multiplicative function of n - see Tóth).
%F Multiplicative: a(p^k) = ( p^(k+2)*(p^2 + 4*p + 1) - p^3*(k + 2)^3 + p^2*(3*k^3 + 15*k^2 + 21*k + 5) - p*(3*k^3 + 12*k^2 + 12*k + 4) + (k + 1)^3 ) / (p - 1)^4. (End)
%t a[n_] := n * DivisorSum[n, DivisorSigma[0, #]^3/# &]; Array[a, 61] (* _Amiram Eldar_, May 09 2021 *)
%o (PARI) a(n) = n*sumdiv(n, d, numdiv(d)^3/d);
%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k)^3*x^k/(1-x^k)^2))
%Y Cf. A007429, A013661, A062369, A097988, A344043.
%K nonn,mult,easy
%O 1,2
%A _Seiichi Manyama_, May 09 2021
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