A308084 - OEIS

%I #59 Nov 05 2025 15:22:41

%S 0,1,3,9,10,30,21,56,54,90,55,198,78,182,210,300,136,459,171,570,420,

%T 462,253,1104,450,650,702,1134,406,1740,465,1488,1056,1122,1190,2835,

%U 666,1406,1482,3120,820,3444,903,2838,2970,2070,1081,5640,1764,3675,2550

%N a(n) = n*(n-1)*d(n)/4, where d(n)=A000005(n) is the number of divisors of n.

%C One eighth of row sums of A308805.

%C When n is not a square, d(n) is even; when n=k^2 for some k, n*(n-1)=(k-1)*k^2*(k+1) is a multiple of 4; in all cases a(n) is an integer.

%H Luc Rousseau, <a href="/A308084/b308084.txt">Table of n, a(n) for n = 1..10000</a>

%H Joerg Arndt, <a href="https://arxiv.org/abs/1202.6525">On computing the generalized Lambert series</a>, arXiv:1202.6525v3 [math.CA], (2012).

%F a(n) = A000217(n-1)*A000005(n)/2.

%F From _Peter Bala_, Jan 21 2021: (Start)

%F G.f.: A(q) = (1/4)*Sum_{n >= 1} q^n*((n*2 + n)*q^n + n^2 - n)/(1 - q^n)^3.

%F Faster converging g.f.: A(q) = (1/4)*Sum_{n >= 1} q^(n^2)*( (n^2 + n)*q^n + n^2 - n)*( (n^2 - n)*q^(2*n) - 2*(n^2 - 1)*q^n + n^2 + n )/(1 - q^n)^3 - differentiate equation 5 in Arndt twice w.r.t. q and set x = 1. (End)

%t Table[n * (n - 1) * DivisorSigma[0, n] / 4, {n, 1, 50}] (* _Amiram Eldar_, Aug 14 2019 *)

%o (PARI)

%o for(n=1,80,print1(n*(n-1)*numdiv(n)/4,", "))

%Y Cf. A000005, A000217, A308805.

%K nonn,easy

%O 1,3

%A _Luc Rousseau_, Aug 07 2019