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.

A007858 G.f. is 1 - 1/f(x), where f(x) = 1+x+3*x^2+9*x^3+32*x^4+... is 1/x times g.f. for A063020.

Original entry on oeis.org

1, 2, 4, 13, 44, 164, 636, 2559, 10556, 44440, 190112, 824135, 3612244, 15981632, 71277736, 320121747, 1446537564, 6571858168, 30000766128, 137544893940, 633051803120, 2923867281660, 13547594977500, 62955434735505, 293336372858724, 1370149533359784, 6414423856436816
Offset: 1

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Author

Martin Klazar, Mar 15 1996

Keywords

Comments

Number of maximal independent sets in rooted plane trees on n nodes. - Olivier Gérard, Jul 05 2001

Crossrefs

Cf. A000108.

Programs

  • Maple
    series(1-x/RootOf(Z-_Z^2-_Z^3+_Z^4-x), x=0,20); # _Mark van Hoeij, May 28 2013
  • Mathematica
    Rest[CoefficientList[1-x/InverseSeries[Series[x-x^2-x^3+x^4, {x, 0, 20}], x],x]] (* Vaclav Kotesovec, Nov 14 2014 *)
    Table[Sum[Binomial[n + k, k]/(n + k)*Sum[(Binomial[j, n - k - j + 1]*Binomial[k, j]*(-1)^(n + k - j + 1)), {j, 0, k}], {k, 1, n}] + CatalanNumber[n], {n, 0, 50}] (* G. C. Greubel, Feb 15 2017 *)
  • Maxima
    a(n):=sum(binomial(n+k,k)/(n+k)*sum(binomial(j,n-k-j+1)*binomial(k,j)*(-1)^(n+k-j+1),j,0,k),k,1,n)+1/(n+1)*binomial(2*n,n); /* Vladimir Kruchinin, Nov 13 2014 */
  • PARI
    my(x='x+O('x^66)); Vec(1-x/serreverse(x-x^2-x^3+x^4)) \\ Joerg Arndt, May 28 2013
    

Formula

a(n+1) = Sum_{k = 1..n} ( binomial(n+k,k)/(n+k)*Sum_{j = 0..k} ( binomial(j,n-k-j+1)*binomial(k,j)*(-1)^(n+k-j+1) ) ) + C(n), where C(n) is a Catalan number. - Vladimir Kruchinin, Nov 13 2014
Recurrence: 16*(n-1)*n*(2*n-3)*(17*n^2 - 81*n + 96)*a(n) = (n-1)*(1819*n^4 - 14124*n^3 + 40377*n^2 - 50320*n + 23040)*a(n-1) + 8*(2*n-5)*(4*n-11)*(4*n-9)*(17*n^2 - 47*n + 32)*a(n-2). - Vaclav Kotesovec, Nov 14 2014
Asymptotics (Klazar, 1997): a(n) ~ sqrt(5731-4635/sqrt(17)) * ((107+51*sqrt(17))/64)^n / (256 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 14 2014