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.

A239112 Nonzero coefficients of g.f. A(x) = 1 + 4*x^2 + 36*x^4 + 396*x^6 + ... satisfying (A-1)*(1+3/A)^3 = 256*x^2.

Original entry on oeis.org

1, 4, 36, 396, 4788, 61200, 811548, 11047176, 153362196, 2161860912, 30850416144, 444692531052, 6464058125868, 94633260282288, 1393927177591080, 20641691487933648, 307096967638064340, 4587717402640280304, 68788081908126980784, 1034809142155901808624
Offset: 0

Views

Author

N. J. A. Sloane, Mar 29 2014

Keywords

Programs

  • Mathematica
    CoefficientList[Series[-2 + 64*x + Sqrt[1 - 256*x + 4096*x^2 + 12*(x - 16*x^2)^(1/3)] + (1/2)*Sqrt[-24 + 32*(1 - 32*x)^2 - 48*(x - 16*x^2)^(1/3) + (8*(1 + 480*x - 24576*x^2 + 262144*x^3)) / Sqrt[1 - 256*x + 4096*x^2 + 12*(x - 16*x^2)^(1/3)]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 30 2014 *)
    Flatten[{1,RecurrenceTable[{256 (-3+n) (-5+3 n) (-1+3 n) a[-2+n]-8 (-15+82 n-108 n^2+36 n^3) a[-1+n]+n (-2+3 n) (-1+3 n) a[n]==0, a[1]==4, a[2]==36}, a, {n,20}]}] (* Vaclav Kotesovec, Mar 30 2014 *)
    terms = 20; B[] = 1; Do[B[x] = 1+256*x/(1+3/B[x])^3 + O[x]^terms, terms]; CoefficientList[B[x], x] (* Jean-François Alcover, Jan 15 2018, after Vaclav Kotesovec *)
  • PARI
    {a(n)=local(B=1+x*O(x^n)); for(i=0, n, B=1+256*x/(1+3/B)^3); polcoeff(B, n)}
    for(n=0, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Mar 30 2014

Formula

From Vaclav Kotesovec, Mar 30 2014: (Start)
G.f. B(x) satisfies: B(x) = 1+256*x/(1+3/B(x))^3.
G.f.: -2 + 64*x + sqrt(1 - 256*x + 4096*x^2 + 12*(x - 16*x^2)^(1/3)) + sqrt(-24 + 32*(1 - 32*x)^2 - 48*(x - 16*x^2)^(1/3) + (8*(1 + 480*x - 24576*x^2 + 262144*x^3))/sqrt(1 - 256*x + 4096*x^2 + 12*(x - 16*x^2)^(1/3)))/2.
D-finite with recurrence n*(3*n-2)*(3*n-1)*a(n) = 8*(36*n^3 - 108*n^2 + 82*n - 15)*a(n-1) - 256*(n-3)*(3*n-5)*(3*n-1)*a(n-2).
a(n) ~ sqrt(3) * GAMMA(1/3) * 2^(4*n-1/3) / (Pi * n^(4/3)) * (1 - 3*2^(1/3) * sqrt(Pi) / (n^(1/3) * GAMMA(1/6))).
(End)

Extensions

a(7)-a(20) from Vaclav Kotesovec, Mar 30 2014