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.

A213589 Antidiagonal sums of the convolution array A213587.

Original entry on oeis.org

1, 6, 20, 55, 135, 308, 668, 1395, 2830, 5610, 10914, 20904, 39515, 73860, 136720, 250937, 457137, 827260, 1488190, 2662905, 4741946, 8407236, 14846100, 26120400, 45801925, 80064018, 139553708, 242597035, 420678315, 727792580
Offset: 1

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Author

Clark Kimberling, Jun 19 2012

Keywords

Crossrefs

Programs

  • GAP
    F:=Fibonacci;; List([1..35], n-> (n+1)*((n+2)*F(n+3) + 2*(n-2)*F(n+2))/10) # G. C. Greubel, Jul 08 2019
  • Magma
    F:=Fibonacci; [(n+1)*((n+2)*F(n+3) + 2*(n-2)*F(n+2))/10: n in [1..35]]; // G. C. Greubel, Jul 08 2019
    
  • Mathematica
    (* First program *)
    b[n_]:= Fibonacci[n+1]; c[n_]:= Fibonacci[n+1];
    T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
    TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
    Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]] (* A213587 *)
    r[n_]:= Table[T[n, k], {k, 40}]  (* columns of antidiagonal triangle *)
    Table[T[n, n], {n, 1, 40}] (* A213588 *)
    s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
    Table[s[n], {n, 1, 50}] (* A213589 *)
    (* Second program *)
    Table[(n+1)*(n*LucasL[n+3] -2*Fibonacci[n])/10, {n, 35}] (* G. C. Greubel, Jul 08 2019 *)
  • Maxima
    a(n):=(n+1)/2*sum((n-j)*binomial(n-j+1,j),j,0,(n+1)/2); /* Vladimir Kruchinin, Apr 09 2016 */
    
  • PARI
    vector(35, n, f=fibonacci; (n+1)*((n+2)*f(n+3)+ 2*(n-2)*f(n+2) )/10) \\ G. C. Greubel, Jul 08 2019
    
  • Sage
    f=fibonacci; [(n+1)*((n+2)*f(n+3)+ 2*(n-2)*f(n+2) )/10 for n in (1..35)] # G. C. Greubel, Jul 08 2019
    

Formula

a(n) = 3*a(n-1) - 5*a(n-3) + 3*a(n-5) + a(n-6).
G.f.: x*(1 + 3*x + 2*x^2)/(1 - x - x^2)^3.
a(n) = (n+1)/2*Sum_{j=0..(n+1)/2}((n-j)*binomial(n-j+1,j)). - Vladimir Kruchinin, Apr 09 2016
a(n) = (n+1)*(n*Lucas(n+3) - 2*Fibonacci(n))/10 = (n+1)*((n+2) *Fibonacci(n+3) + 2*(n-2)*Fibonacci(n+2))/10. - G. C. Greubel, Jul 08 2019