A122445 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + 2*T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0) = T(n+1,2n) = 1 and T(n+1,2n+1) = T(n+1,2n+2) = 0.

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 10, 8, 3, 1, 0, 0, 1, 4, 10, 22, 36, 28, 12, 4, 1, 0, 0, 1, 5, 15, 39, 83, 135, 107, 47, 17, 5, 1, 0, 0, 1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0, 1, 7, 28, 92, 259, 629, 1298, 2094, 1676, 773, 305, 104, 30, 7, 1, 0, 0

Offset: 0

Author

Paul D. Hanna, Sep 07 2006

Keywords

• nonn
• tabf

Comments

The diagonals may be generated by iterated convolutions of a base sequence B with the sequence C of central terms. The g.f. B(x) of the base sequence satisfies: B = 1 + x*B^2 + 2x^2*(B^2 - B); the g.f. C(x) of the central terms satisfies: C(x) = 1/(1+x - xB(x)).

Examples

			To obtain row 4, pendular sums of row 3 are carried out as follows.
  [1, 2, 3,  2, 1, 0, 0]: given row 3;
  [1, _, _, __, _, _, _]: start with T(4,0) = T(3,0) = 1;
  [1, _, _, __, _, _, 1]: T(4,6) = T(4,0) + 2*T(3,6) = 1 + 2*0 = 1;
  [1, 3, _, __, _, _, 1]: T(4,1) = T(4,6) + 1*T(3,1) = 1 + 1*2 = 3;
  [1, 3, _, __, _, 3, 1]: T(4,5) = T(4,1) + 2*T(3,5) = 3 + 2*0 = 3;
  [1, 3, 6, __, _, 3, 1]: T(4,2) = T(4,5) + 1*T(3,2) = 3 + 1*3 = 6;
  [1, 3, 6, __, 8, 3, 1]: T(4,4) = T(4,2) + 2*T(3,4) = 6 + 2*1 = 8;
  [1, 3, 6, 10, 8, 3, 1]: T(4,3) = T(4,4) + 1*T(3,3) = 8 + 1*2 = 10;
  [1, 3, 6, 10, 8, 3, 1,0,0]: complete row 4 by appending two zeros.
Triangle begins:
  1;
  1, 0,  0;
  1, 1,  1,  0,   0;
  1, 2,  3,  2,   1,   0,   0;
  1, 3,  6, 10,   8,   3,   1,   0,   0;
  1, 4, 10, 22,  36,  28,  12,   4,   1,  0,  0;
  1, 5, 15, 39,  83, 135, 107,  47,  17,  5,  1, 0, 0;
  1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0;
Central terms are:
  C = A122447 = [1, 0, 1, 2, 8, 28, 107, 418, 1676, 6848, ...].
Lower diagonals start:
  D1 = A122448 = [1, 1, 3, 10, 36, 135, 525, 2094, 8524, ...];
  D2 = A122449 = [1, 2, 6, 22, 83, 324, 1298, 5302, 22002, ...].
Diagonals above central terms (ignoring leading zeros) start:
  U1 = A122450 = [1, 3, 12, 47, 189, 773, 3208, 13478, 57222, ...];
  U2 = A122451 = [1, 4, 17, 72, 305, 1300, 5576, 24068, 104510, ...].
There exists the base sequence:
  B = A122446 = [1, 1, 2, 7, 24, 88, 336, 1321, 5316, 21788, ...]
which generates all diagonals by convolutions with central terms:
  D2 = B * D1 = B^2 * C
  U2 = B * U1 = B^2 * C"
where C" = [1, 2, 8, 28, 107, 418, 1676, 6848, 28418, ...]
are central terms not including the initial [1,0].
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A122446, A122447 (central terms), A122452 (row sums).

Diagonals: A122448, A122449, A122450, A122451.

Variants: A118340, A118345, A118350, A119369.

Programs

Maple

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k<0 or k>2*(n-1) then 0
    elif n=2 and k<3 then 1
    else T(n-1, k) + `if`(kG. C. Greubel, Mar 16 2021

Mathematica

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, T[n-1, k] + If[kG. C. Greubel, Mar 16 2021 *)

PARI

{T(n,k)= if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, if(k
				

Sage

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    else: return T(n-1, k) + ( T(n, 2*n-k-1) if kG. C. Greubel, Mar 16 2021