A278403 a(n) = Sum_{d|n} d^2 * (d+1)/2.

1, 7, 19, 47, 76, 151, 197, 335, 424, 632, 727, 1127, 1184, 1673, 1894, 2511, 2602, 3634, 3611, 4872, 5066, 6299, 6349, 8615, 8201, 10316, 10630, 13081, 12616, 16526, 15377, 19407, 19258, 22838, 22322, 28586, 26012, 31775, 31622, 37960, 35302, 44594, 40679, 49899, 48874, 56081, 53017, 67239, 60222, 72507, 70246, 82012, 75844, 94030, 85502, 102745, 97850, 111860, 104431, 131502

Offset: 1

Author

Paul D. Hanna, Nov 20 2016

Keywords

• nonn
• easy

Examples

			L.g.f.: L(x) = x + 7*x^2/2 + 19*x^3/3 + 47*x^4/4 + 76*x^5/5 + 151*x^6/6 + 197*x^7/7 + 335*x^8/8 + 424*x^9/9 + 632*x^10/10 + 727*x^11/11 + 1127*x^12/12 +...
which equals the series
L(x) = x/(1-x)^3 + (x^2/2)/(1-x^2)^3 + (x^3/3)/(1-x^3)^3 + (x^4/4)/(1-x^4)^3 + (x^5/5)/(1-x^5)^3 + (x^6/6)/(1-x^6)^3 + (x^7/7)/(1-x^7)^3 +...
The exponentiation of the l.g.f. equals the infinite product
exp(L(x)) = 1/((1-x)*(1-x^2)^3*(1-x^3)^6*(1-x^4)^10*(1-x^5)^15*(1-x^6)^21*...);
explicitly,
exp(L(x)) = 1 + x + 4*x^2 + 10*x^3 + 26*x^4 + 59*x^5 + 141*x^6 + 310*x^7 + 692*x^8 + 1483*x^9 + 3162*x^10 + 6583*x^11 + 13602*x^12 +...+ A000294(n)*x^n +...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000294, A001157, A001158, A277613.

Programs

Mathematica

Table[Total[#^2*(#+1)/2&/@Divisors[n]],{n,60}] (* Harvey P. Dale, Jul 26 2017 *)

PARI

{a(n) = sumdiv(n,d,d^2*(d+1)/2)}
for(n=1,60,print1(a(n),", "))

PARI

{a(n) = (sigma(n,3) + sigma(n,2))/2}
for(n=1,60,print1(a(n),", "))

PARI

{a(n) = n * polcoeff( sum(k=1,n, (x^k/k) / (1 - x^k +x*O(x^n))^3), n)}
for(n=1,60,print1(a(n),", "))

Formula

Let the l.g.f. be L(x) = Sum_{n>=1} a(n)*x^n/n, then:

(1) exp( L(x) ) = Product_{n>=1} 1/(1 - x^n)^(n*(n+1)/2),

(2) L(x) = Sum_{n>=1} (x^n/n) / (1 - x^n)^3.

O.g.f.: Sum_{n>=1} n^2*(n+1)/2 * x^n / (1 - x^n).

a(n) = (sigma_3(n) + sigma_2(n))/2, where sigma_2(n) = A001157(n) and sigma_3(n) = A001158(n).

Sum_{k=1..n} a(k) ~ Pi^4 * n^4 / 720. - Vaclav Kotesovec, Jul 13 2021

Dirichlet g.f.: zeta(s) * (zeta(s-3) + zeta(s-2)) / 2. - Amiram Eldar, Jan 02 2025