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.

A093951 Sum of integers generated by n-1 substitutions, starting with 1, k -> k+1, k-1, .., 1.

Original entry on oeis.org

1, 2, 4, 8, 17, 36, 80, 176, 403, 910, 2128, 4896, 11628, 27132, 65208, 153824, 373175, 888030, 2170740, 5202600, 12797265, 30853680, 76292736, 184863168, 459162452, 1117370696, 2786017120, 6804995008, 17024247304, 41717833740, 104673837384
Offset: 1

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Author

Wouter Meeussen, Apr 18 2004

Keywords

Comments

Substitutions 1 -> {2}, 2 -> {3,1}, 3 -> {4,2}, 4 -> {5,3,1}, 5 -> {6,4,2}, 6 -> {7,5,3,1}, 7 -> {8,6,4,2}, etc. The function f(n) gives determinant of (I_{n} - x * A(n)) where I_{n} is the identity matrix and A(n) = 0 if j > i + 1 otherwise (i+j) mod 2, for i = 1, ..., n and j = 1, ..., n, and can be written in terms of Dickson polynomials as g(w) = x*D_(w-1)(1+x, x*(1+x)) + (1-2*x)*E_(w-1)(1+x, x*(1+x)). - Francisco Salinas (franciscodesalinas(AT)hotmail.com), Apr 13 2004
Count of integers is A047749.
Sum of integers with substitution starting from 0 is A084081.

Examples

			GF(12) = (1 + 2*x - 7*x^2 - 14*x^3 + 9*x^4 + 20*x^5 + 2*x^6 - 2*x^7 + 2*x^11)/(1 - 11*x^2 + 36*x^4 - 35*x^6 + 5*x^8) produces a(1) to a(12).
a(4)=8 since 4-1 = 3 substitutions on 1 produce 1 -> 2 -> 3+1 -> 4 + 2 + 2 = 8.
		

Crossrefs

Programs

  • Magma
    function A093951(n)
      if (n mod 2) eq 0 then return 8*Binomial(Floor(3*n/2), Floor((n-2)/2))/(n+2);
      else return 6*Binomial(Floor((3*n+1)/2), Floor((n-1)/2))/(n+2) - 2*Binomial(Floor((3*n-1)/2), Floor((n-1)/2))/(n+1);
      end if; return A093951;
    end function;
    [A093951(n): n in [1..40]]; // G. C. Greubel, Oct 17 2022
    
  • Mathematica
    Plus@@@Flatten/@NestList[ #/.k_Integer:>Range[k+1, 1, -2]&, {1}, 8];(*or for n>16 *); f[1]=1; f[2]=1-x^2; f[3]=1-2x^2; f[n_]:=f[n]=Expand[f[n-1]-x^2 f[n-3]]; g[1]=1; g[2]=1+2x; g[3]=1+2x+2x^2; g[n_]:=g[n]=Expand[g[n-1] -x^2 g[n-3]+2 x^(n-1)]; GF[n_]:=g[n]/f[n]; CoefficientList[Series[GF[36], {x, 0, 36}], x]
  • PARI
    {a(n)=if(n%2==0,4*binomial(3*n/2,n/2-1)/(n/2+1), 6*binomial(3*(n\2)+2, n\2)/(2*(n\2)+3) - binomial(3*(n\2)+1,n\2)/(n\2+1))} \\ Paul D. Hanna, Apr 24 2006
    
  • SageMath
    def A093951(n):
        if (n%2==0): return 8*binomial(3*n/2, (n-2)/2)/(n+2)
        else: return 6*binomial((3*n+1)/2, (n-1)/2)/(n+2) - 2*binomial((3*n-1)/2, (n-1)/2)/(n+1)
    [A093951(n) for n in range(1,40)] # G. C. Greubel, Oct 17 2022

Formula

a(n) = [x^n] GF(n) with GF(n) = g(n)/f(n) and f(1)=1, f(2)=1-x^2, f(3)=1-2*x^2, f(n) = f(n-1) - x^2*f(n-3) and g(1)=1, g(2)=1+2*x, g(3)=1+2*x+2*x^2, g(n) = g(n-1) - x^2*g(n-3) + 2*x^(n-1).
From Paul D. Hanna, Apr 24 2006: (Start)
a(2*n) = 4*binomial(3*n, n-1)/(n+1) = 2*A006629(n-1).
a(2*n+1) = 6*binomial(3*n+2, n)/(2*n+3) - binomial(3*n+1, n)/(n+1) = A056096(n+3). (End)