A135299 Pascal's triangle, but the last element of the row is the sum of the 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

0,3

Links

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

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).

Example

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)

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 *)

Crossrefs

Cf. A007318, A067337.

Sequence in context: A019224 A298096 A053190 * A092081 A203997 A057740

Adjacent sequences: A135296 A135297 A135298 * A135300 A135301 A135302

Keyword

nonn,tabl

Author

Jose Ramon Real, Dec 04 2007

Status

approved