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.

A084075 Length of list created by n substitutions k -> Range( -abs(k+1), abs(k-1), 2) starting with {1}.

Original entry on oeis.org

1, 2, 5, 12, 33, 86, 249, 680, 2033, 5722, 17485, 50260, 156033, 455534, 1431281, 4228752, 13412193, 40003058, 127840085, 384232156, 1235575201, 3737280582, 12080678505, 36736735672, 119276490193, 364372758986, 1187542872989
Offset: 0

Views

Author

Wouter Meeussen, May 11 2003

Keywords

Examples

			{1}, {-2,0}, {-1,1,3,-1,1}, {0,2,-2,0,-4,-2,0,2,0,2,-2,0}
		

Crossrefs

Cf. A027307, A215067, A034015 (even bisection).

Programs

  • Magma
    I:=[1,2,5,12]; [n le 4 select I[n] else (6*(35*n^2-55*n-76)*Self(n-1) + (275*n^4-770*n^3-203*n^2+1736*n-912)*Self(n-2) -6*(5*n^2+5*n-28)*Self(n-3) + (n-4)*(n-2)*(25*n^2+5*n-48)*Self(n-4))/(n*(n+2)*(25*n^2-45*n-28)): n in [1..41]]; // G. C. Greubel, Nov 24 2022
    
  • Mathematica
    Rest@CoefficientList[InverseSeries[Series[ (-1-6n-8n^2+(1+2n)^2 Sqrt[1+4n])/( 2(n+4n^2+4n^3)), {n, 0, 40}]], n]
    Length/@Flatten/@NestList[ #/.k_Integer:>Range[-Abs[k+1], Abs[k-1], 2] &, {1}, 8]
  • Python
    # replace iterates lists as described in Example.
    def replace(L):
        return [i for k in L for i in range(-abs(k + 1), 1 + abs(k - 1), 2)]
    def a(n):
      L = [1]
      for k in range(n): L=replace(L)
      return len(L)
    print([a(n) for n in range(12)]) # F. Chapoton, Nov 15 2024
  • SageMath
    @CachedFunction
    def a(n): # a = A084075
        if n < 4: return (1, 2, 5, 12)[n]
        else: return (6*(35*n^2 +15*n -96)*a(n-1) +(275*n^4+330*n^3-863*n^2+120*n+126)*a(n-2) -6*(5*n^2+15*n-18)*a(n-3) +(n-3)*(n-1)*(25*n^2+55*n-18)*a(n-4))/((n+1)*(n+3)*(25*n^2+5*n-48))
    [a(n) for n in range(41)] # G. C. Greubel, Nov 24 2022
    

Formula

G.f. is the series reversion of (-1 -6*x -8*x^2 + (1+2*x)^2 * sqrt(1+4*x))/(2*(x +4*x^2 +4*x^3)).
a(2*n) = A027307(n)/2, n >= 1.
a(n) = ( 6*(35*n^2 +15*n -96)*a(n-1) + (275*n^4 +330*n^3 -863*n^2 +120*n +126)*a(n-2) - 6*(5*n^2 +15*n -18)*a(n-3) + (n-3)*(n-1)*(25*n^2 +55*n -18)*a(n-4) )/((n+1)*(n+3)*(25*n^2 +5*n -48)), n >= 4. - G. C. Greubel, Nov 24 2022