A356692 Pascal-like triangle, where each entry is the sum of the four entries above it starting with 1 at the top.

1, 1, 1, 2, 2, 2, 4, 6, 6, 4, 10, 16, 20, 16, 10, 26, 46, 62, 62, 46, 26, 72, 134, 196, 216, 196, 134, 72, 206, 402, 618, 742, 742, 618, 402, 206, 608, 1226, 1968, 2504, 2720, 2504, 1968, 1226, 608, 1834, 3802, 6306, 8418, 9696, 9696, 8418, 6306, 3802, 1834, 5636, 11942, 20360, 28222, 34116, 36228, 34116, 28222, 20360, 11942, 5636

Offset: 0

Author

Greg Dresden and Sadek Mohammed, Aug 23 2022

Keywords

• nonn
• tabl

Comments

Similar in spirit to the regular Pascal triangle, except that here we have T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k) + T(n-1,k+1), with the understanding that T(0,0) is defined to be 1, and T(n,k) is defined as 0 for k<0 and k>n.

T(n,k) is the number of permutations p of [n+1] such that at most one element of {p(1),...,p(i-1)} is between p(i) and p(i+1) for all i <= n and p(n+1) = k+1. T(4,1) = 16: 13542, 14532, 15342, 15432, 31542, 35142, 35412, 41532, 45132, 45312, 51342, 51432, 53142, 53412, 54132, 54312. - Alois P. Heinz, Aug 31 2022

Examples

			T(4,0) = 10 because it is the sum of T(3,-2), T(3,-1), T(3,0), and T(3,1) which gives 0+0+4+6 = 10.
Triangle begins:
             1
           1   1
         2   2   2
       4   6   6   4
     10  16  20  16  10
   26  46  62  62  46  26
  ...
		

Links

• Alois P. Heinz, Rows n = 0..150, flattened

Crossrefs

Row sums give A216837(n+1).

Column k=0 and also main diagonal give A356832.

T(2n,n) gives A356853.

Cf. A007318, A064189.

Programs

Maple

T:= proc(n, k) option remember; `if`(k<0 or k>n, 0,
      `if`(n=0, 1, add(T(n-1,j), j=k-2..k+1)))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Aug 28 2022

Mathematica

T[0, 0] = 1; T[n_, k_] := T[n, k] = If[k < 0 || k > n, 0,
    T[n - 1, k - 2] + T[n - 1, k - 1] + T[n - 1, k] + T[n - 1, k + 1]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten

Formula

T(n,k) = T(n,n-k).