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.

A118345 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + 2*T(n-1,k), for n>=k>=0, with T(n,0) = 1 and T(n,n) = 0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 5, 1, 0, 1, 4, 11, 6, 1, 0, 1, 5, 18, 30, 7, 1, 0, 1, 6, 26, 70, 40, 8, 1, 0, 1, 7, 35, 121, 201, 51, 9, 1, 0, 1, 8, 45, 184, 487, 286, 63, 10, 1, 0, 1, 9, 56, 260, 873, 1445, 386, 76, 11, 1, 0, 1, 10, 68, 350, 1375, 3592, 2147, 502, 90, 12, 1, 0
Offset: 0

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Author

Paul D. Hanna, Apr 26 2006

Keywords

Comments

See A118340 for definition of pendular triangles and pendular sums.

Examples

			Row 6 equals the pendular sums of row 5:
  [1,  4, 11,  6,  1,  0], where the pendular sums proceed as follows:
  [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1]: T(6,5) = T(6,0) + 2*T(5,5) = 1 + 2*0 = 1;
  [1,  5, __, __, __,  1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  7,  1]: T(6,4) = T(6,1) + 2*T(5,4) = 5 + 2*1 = 7;
  [1,  5, 18, __,  7,  1]: T(6,2) = T(6,4) + T(5,2) = 7 + 11 = 18;
  [1,  5, 18, 30,  7,  1]: T(6,3) = T(6,2) + 2*T(5,3) = 18 + 2*6 = 30;
  [1,  5, 18, 30,  7,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  5,   1,    0;
  1,  4, 11,   6,    1,    0;
  1,  5, 18,  30,    7,    1,    0;
  1,  6, 26,  70,   40,    8,    1,   0;
  1,  7, 35, 121,  201,   51,    9,   1,  0;
  1,  8, 45, 184,  487,  286,   63,  10,  1,  0;
  1,  9, 56, 260,  873, 1445,  386,  76, 11,  1, 0;
  1, 10, 68, 350, 1375, 3592, 2147, 502, 90, 12, 1, 0; ...
Central terms are T(2*n,n) = A118346(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118347(n) = [A118346^2](n),
T(2*n+2,n) = A118348(n) = [A118346^3](n).
		

Crossrefs

Cf. A167763 (p=0), A118340 (p=1), this sequence (p=2), A118350 (p=3).

Programs

  • Magma
    function T(n,k,p)
      if k lt 0 or n lt k then return 0;
      elif k eq 0 then return 1;
      elif k eq n then return 0;
      elif n gt 2*k then return T(n,n-k,p) + T(n-1,k,p);
      else return T(n,n-k-1,p) + p*T(n-1,k,p);
      end if;
      return T;
    end function;
    [T(n,k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2021
  • Mathematica
    T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 17 2021 *)
  • PARI
    T(n,k)=if(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)+2*T(n-1,k)))))
    
  • Sage
    @CachedFunction
    def T(n, k, p):
        if (k<0 or n2*k): return T(n,n-k,p) + T(n-1,k,p)
        else: return T(n, n-k-1, p) + p*T(n-1, k, p)
    flatten([[T(n,k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021
    

Formula

T(2*n+m,n) = [A118346^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A118346.