A119369 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) + 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, 9, 7, 3, 1, 0, 0, 1, 4, 10, 20, 30, 23, 11, 4, 1, 0, 0, 1, 5, 15, 36, 70, 104, 81, 40, 16, 5, 1, 0, 0, 1, 6, 21, 58, 133, 253, 374, 293, 149, 63, 22, 6, 1, 0, 0, 1, 7, 28, 87, 226, 501, 938, 1380, 1087, 564, 248, 93, 29, 7, 1, 0, 0

Offset

0,11

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 + x^2*(B^2 - B); the g.f. C(x) of the central terms satisfies: C(x) = 1/(1+x - x*B(x)).

Links

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

Formula

Sum_{k=0..2*n} T(n, k) = A119372(n). - G. C. Greubel, Mar 16 2021

Example

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) + T(3,6) = 1 + 0 = 1;
  [1, 3, _, _, _, _, 1]: T(4,1) = T(4,6) + T(3,1) = 1 + 2 = 3;
  [1, 3, _, _, _, 3, 1]: T(4,5) = T(4,1) + T(3,5) = 3 + 0 = 3;
  [1, 3, 6, _, _, 3, 1]: T(4,2) = T(4,5) + T(3,2) = 3 + 3 = 6;
  [1, 3, 6, _, 7, 3, 1]: T(4,4) = T(4,2) + T(3,4) = 6 + 1 = 7;
  [1, 3, 6, 9, 7, 3, 1]: T(4,3) = T(4,4) + T(3,3) = 7 + 2 = 9;
  [1, 3, 6, 9, 7, 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,  9,   7,   3,   1,    0,    0;
  1, 4, 10, 20,  30,  23,  11,    4,    1,   0,   0;
  1, 5, 15, 36,  70, 104,  81,   40,   16,   5,   1,  0,  0;
  1, 6, 21, 58, 133, 253, 374,  293,  149,  63,  22,  6,  1, 0, 0;
  1, 7, 28, 87, 226, 501, 938, 1380, 1087, 564, 248, 93, 29, 7, 1, 0, 0;
Central terms are:
  C = A119371 = [1, 0, 1, 2, 7, 23, 81, 293, 1087, 4110, ...].
Lower diagonals start:
  D1 = A119372 = [1, 1, 3, 9, 30, 104, 374, 1380, 5197, ...];
  D2 = A119373 = [1, 2, 6, 20, 70, 253, 938, 3546, 13617, ...].
  Diagonals above central terms (ignoring leading zeros) start:
  U1 = A119375 = [1, 3, 11, 40, 149, 564, 2166, 8420, ...];
  U2 = A119376 = [1, 4, 16, 63, 248, 980, 3894, 15563, ...].
There exists the base sequence:
  B = A119370 = [1, 1, 2, 6, 19, 64, 225, 816, 3031, 11473, ...]
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, 7, 23, 81, 293, 1087, ...]
are central terms not including the initial [1,0].

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`(k<n, T(n, 2*n-k-1), T(n, 2*n-k-2))
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. 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[k<n, T[n, 2*n-k-1], T[n, 2*n-k-2] ]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, 2*n}]//Flatten (* G. C. Greubel, Mar 16 2021 *)

Prog

(PARI) T(n, k)= if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, T(n-1, k) + if(k<n, T(n, 2*n-1-k), T(n, 2*n-2-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 k<n else T(n, 2*n-k-2) )
flatten([[T(n, k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 16 2021

Crossrefs

Cf. A119370, A119371, A119372, A119373, A119374, A119375, A119376.

Variants: A118340, A118345, A118350.

Sequence in context: A178904 A017858 A167769 * A122445 A189511 A165592

Adjacent sequences: A119366 A119367 A119368 * A119370 A119371 A119372

Keyword

nonn,tabf

Author

Paul D. Hanna, May 16 2006

Status

approved