A135299 Pascal's triangle, but the last element of the row is the sum of all the previous terms.

1, 1, 2, 1, 3, 8, 1, 4, 11, 32, 1, 5, 15, 43, 128, 1, 6, 20, 58, 171, 512, 1, 7, 26, 78, 229, 683, 2048, 1, 8, 33, 104, 307, 912, 2731, 8192, 1, 9, 41, 137, 411, 1219, 3643, 10923, 32768, 1, 10, 50, 178, 548, 1630, 4862, 14566, 43691, 131072

Offset: 0

Author

Jose Ramon Real, Dec 04 2007

Keywords

• nonn
• tabl

Examples

			T(2,1) = T(1,0) + T(1,1) = 1 + 2 = 3;
T(2,2) = T(0,0) + T(1,0) + T(1,1) + T(2,0) + T(2,1) = 1 + 1 + 2 + 1 + 3 = 8.
From _G. C. Greubel_, Oct 09 2016: (Start)
The triangle is:
  1;
  1, 2;
  1, 3,  8;
  1, 4, 11, 32;
  1, 5, 15, 43, 128;
  1, 6, 20, 58, 171, 512;
  ... (End)
		

Links

• G. C. Greubel, Table of n, a(n) for the first 25 rows

Crossrefs

Cf. A007318, A067337.

Programs

Mathematica

T[0, 0] := 1; T[n_, 0] := 1; T[n_, k_] := T[n - 1, k] + T[n - 1, k - 1]; T[n_, n_] := 2^(2*n - 1); Table[T[n, k], {n, 0, 5}, {k, 0, n}] (* G. C. Greubel, Oct 09 2016 *)

Formula

T(0,0) = 1;

T(n,0) = 1;

T(n,k) = T(n-1, k-1) + T(n-1, k) if k < n;

T(n,n) = (Sum_{j=0..n-1} Sum_{i=0..j} T(j,i)) + Sum_{i=0..n-1} T(n,i) [i.e., sum of all earlier terms of the triangle].

T(n,n) = (4^n)/2 for n > 0;

T(n,n) = 2*Sum_{i=0..n-1} T(n,i).