A185003 a(n) = Sum_{k=1..n} binomial(n,k)*sigma(k).

1, 5, 16, 45, 116, 284, 673, 1557, 3535, 7910, 17502, 38376, 83500, 180479, 387881, 829605, 1766998, 3749765, 7931114, 16724870, 35173777, 73794660, 154485527, 322771344, 673155141, 1401536934, 2913490375, 6047714599, 12536770558, 25956242579, 53678385266

Offset: 1

Author

Paul D. Hanna, Feb 04 2012

Keywords

• nonn

Examples

			L.g.f.: L(x) = x + 5*x^2/2 + 16*x^3/3 + 45*x^4/4 + 116*x^5/5 + 284*x^6/6 +...
where exponentiation yields A103446 (with offset=0):
exp(L(x)) = 1 + x + 3*x^2 + 8*x^3 + 21*x^4 + 54*x^5 + 137*x^6 + 344*x^7 +...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A103446, A222115, A000203, A160399.

Programs

Magma

[&+[Binomial(n,k)*DivisorSigma(1,k):k in [1..n]]:n in [1..31]]; // Marius A. Burtea, Nov 12 2019

Magma

[&+[&+[i*Binomial(n,i*j):j in [1..n]]:i in [1..n]]:n in [1..31]]; // Marius A. Burtea, Nov 12 2019

Maple

with(numtheory): seq(add(binomial(n,i)*sigma(i), i=1..n), n=1..40); # Ridouane Oudra, Nov 12 2019

Mathematica

Table[Sum[Binomial[n, k] DivisorSigma[1, k], {k, n}], {n,50}] (* G. C. Greubel, Jun 03 2017 *)

PARI

{a(n)=sum(k=1,n,sigma(k)*binomial(n,k))}
for(n=1,30,print1(a(n),", "))

PARI

{a(n)=local(X=x+x*O(x^n)); n*polcoeff(sum(m=1, n+1, x^m/((1-x)^m-X^m)/m), n)}

PARI

{a(n)=local(X=x+x*O(x^n)); n*polcoeff(sum(k=1, n, k*log(1-X)-log((1-x)^k-X^k)), n)}

PARI

{a(n)=local(X=x+x*O(x^n)); n*polcoeff(sum(m=1, n+1, sigma(m)*x^m/(1-X)^m/m), n)}

PARI

{a(n)=local(X=x+x*O(x^n)); n*polcoeff(sum(k=1, n, valuation(2*k, 2)*log(1 + x^k/(1-X)^k)), n)}

Formula

Logarithmic derivative of A103446 (with offset=0), which describes the binomial transform of partitions.

From Paul D. Hanna, Jun 01 2013: (Start)

L.g.f.: Sum_{n>=1} sigma(n) * x^n/(1-x)^n / n.

L.g.f.: Sum_{n>=1} x^n/((1-x)^n - x^n) / n.

L.g.f.: Sum_{n>=1} n*log(1-x) - log((1-x)^n - x^n).

L.g.f.: Sum_{n>=1} A001511(n) * log(1 + x^n/(1-x)^n), where 2^A001511(n) is the highest power of 2 that divides 2*n.

a(n) = A222115(n) - 1. (End)

a(n) ~ Pi^2/12 * n * 2^n. - Vaclav Kotesovec, Dec 30 2015

a(n) = Sum_{i=1..n} Sum_{j=1..n} i*binomial(n,i*j). - Ridouane Oudra, Nov 12 2019