cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A222115 a(n) = 1 + Sum_{k=1..n} binomial(n,k) * sigma(k).

Original entry on oeis.org

2, 6, 17, 46, 117, 285, 674, 1558, 3536, 7911, 17503, 38377, 83501, 180480, 387882, 829606, 1766999, 3749766, 7931115, 16724871, 35173778, 73794661, 154485528, 322771345, 673155142, 1401536935, 2913490376, 6047714600, 12536770559, 25956242580, 53678385267, 110889844998
Offset: 1

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Author

Paul D. Hanna, Jun 01 2013

Keywords

Comments

Here sigma(n) is the sum of divisors of n (A000203).

Examples

			L.g.f.: L(x) = 2*x + 6*x^2/2 + 17*x^3/3 + 46*x^4/4 + 117*x^5/5 + 285*x^6/6 +...
where
exp(L(x)) = 1 + 2*x + 5*x^2 + 13*x^3 + 34*x^4 + 88*x^5 + 225*x^6 + 569*x^7 +...+ A218481(n)*x^n +...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]DivisorSigma[1,k],{k,n}],{n,40}]+1 (* Harvey P. Dale, Jul 21 2015 *)
  • PARI
    {a(n)=1+sum(k=1,n,binomial(n,k)*sigma(k))}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(X=x+x*O(x^n)); n*polcoeff(-log(1-X)+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(-log(1-X)+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(-log(1-X)+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(-log(1-X)+sum(k=1, n, valuation(2*k, 2)*log(1 + x^k/(1-X)^k)), n)}

Formula

Logarithmic derivative of the binomial transform of the partition numbers (A218481).
L.g.f.: -log(1-x) + Sum_{n>=1} sigma(n) * x^n/(1-x)^n / n.
L.g.f.: -log(1-x) + Sum_{n>=1} x^n/((1-x)^n - x^n) / n.
L.g.f.: -log(1-x) + Sum_{n>=1} n*log(1-x) - log((1-x)^n - x^n).
L.g.f.: -log(1-x) + 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) = A185003(n) + 1.
a(n) ~ Pi^2/12 * n * 2^n. - Vaclav Kotesovec, Dec 30 2015