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.

Showing 1-2 of 2 results.

A235385 E.g.f.: exp( Sum_{n>=1} H(n) * x^(2*n-1)/(2*n-1) ) where H(n) is the n-th harmonic number.

Original entry on oeis.org

1, 1, 1, 4, 13, 75, 415, 3160, 24545, 233509, 2323165, 26599780, 321545365, 4312503655, 61219938915, 942271981240, 15340303899265, 266671144108265, 4892612440317145, 94840103781865060, 1934826541931748925, 41387703314570495875, 928953515444722956775, 21738929496091877729400
Offset: 0

Views

Author

Paul D. Hanna, Jan 08 2014

Keywords

Comments

Compare to: exp( Sum_{n>=1} x^(2*n-1)/(2*n-1) ) = sqrt(1-x^2)/(1-x).

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 4*x^3/3! + 13*x^4/4! + 75*x^5/5! +...
where
log(A(x)) = x + (1+1/2)*x^3/3 + (1+1/2+1/3)*x^5/5 + (1+1/2+1/3+1/4)*x^7/7 + (1+1/2+1/3+1/4+1/5)*x^9/9 + (1+1/2+1/3+1/4+1/5+1/6)*x^11/11 +...
Explicitly,
log(A(x)) = x + 1/2*x^3 + 11/30*x^5 + 25/84*x^7 + 137/540*x^9 + 49/220*x^11 + 363/1820*x^13 + 761/4200*x^15 +...
Equivalently,
log(A(x)) = x + 3*x^3/3! + 44*x^5/5! + 1500*x^7/7! + 92064*x^9/9! + 8890560*x^11/11! + 1241982720*x^13/13! + 236938141440*x^15/15! +...
		

Crossrefs

Programs

  • PARI
    {H(n)=sum(k=1,n,1/k)}
    {a(n)=local(A=1);A=exp(sum(k=1,n\2+1,H(k)*x^(2*k-1)/(2*k-1))+x*O(x^n));n!*polcoeff(A,n)}
    for(n=0,25,print1(a(n),", "))

A336289 a(0) = 1; a(n) = n! * Sum_{k=1..n} binomial(n-1,k-1) * (k-1)! * H(k) * a(n-k) / (n-k)!, where H(k) is the k-th harmonic number.

Original entry on oeis.org

1, 1, 5, 55, 1054, 31046, 1299386, 73211510, 5338080280, 488727800664, 54865512897432, 7408400404206792, 1184230737883333680, 221121985937352261360, 47683177920267470877648, 11758982455716373002624816, 3287966057434181416523799936
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 16 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = n! Sum[Binomial[n - 1, k - 1] (k - 1)! HarmonicNumber[k] a[n - k]/(n - k)!, {k, 1, n}]; Table[a[n], {n, 0, 16}]
    nmax = 16; CoefficientList[Series[Exp[Sum[HarmonicNumber[k] x^k/k, {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!^2
    nmax = 16; CoefficientList[Series[Exp[Log[1 - x]^2/2 + PolyLog[2, x]], {x, 0, nmax}], x] Range[0, nmax]!^2

Formula

Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(Sum_{n>=1} H(n) * x^n / n).
Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(log(1 - x)^2 / 2 + polylog(2,x)).
Showing 1-2 of 2 results.