A108350 Number triangle T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*((j+1) mod 2).

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 32, 21, 6, 1, 1, 7, 31, 65, 65, 31, 7, 1, 1, 8, 43, 116, 161, 116, 43, 8, 1, 1, 9, 57, 189, 341, 341, 189, 57, 9, 1, 1, 10, 73, 288, 645, 842, 645, 288, 73, 10, 1, 1, 11, 91, 417, 1121, 1827, 1827, 1121, 417, 91

Offset

0,5

Comments

Or as a square array read by antidiagonals, T(n,k) = Sum_{j=0..n} binomial(k,j)*binomial(n+k-j,k)*((j+1) mod 2).

A symmetric number triangle based on 1/(1-x^2).

The construction of a symmetric triangle in this example is general. Let f(n) be a sequence, preferably with f(0)=1. Then T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*f(j) yields a symmetric triangle. When f(n)=1^n, we get Pascal's triangle. When f(n)=2^n, we get the Delannoy triangle (see A008288). In general, f(n)=k^n yields a (1,k,1)-Pascal triangle (see A081577, A081578). Row sums of triangle are A100131. Diagonal sums of the triangle are A108351. Triangle mod 2 is A106465.

Links

Table of n, a(n) for n=0..75.

Formula

Row k (and column k) has g.f. (1+C(k,2)x^2)/(1-x)^(k+1).

Example

Triangle rows begin
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 13, 13,  5,  1;
  1,  6, 21, 32, 21,  6,  1;
As a square array read by antidiagonals, rows begin
  1, 1,  1,   1,   1,    1,    1, ...
  1, 2,  3,   4,   5,    6,    7, ...
  1, 3,  7,  13,  21,   31,   43, ...
  1, 4, 13,  32,  65,  116,  189, ...
  1, 5, 21,  65, 161,  341,  645, ...
  1, 6, 31, 116, 341,  842, 1827, ...
  1, 7, 43, 189, 645, 1827, 4495, ...

Prog

(PARI) trgn(nn) = {for (n= 0, nn, for (k = 0, n, print1(sum(j=0, n-k, binomial(k, j)*binomial(n-j, k)*((j+1) % 2)), ", "); ); print(); ); } \\ Michel Marcus, Sep 11 2013

Crossrefs

Sequence in context: A094525 A130671 A114197 * A086617 A094526 A088699

Adjacent sequences: A108347 A108348 A108349 * A108351 A108352 A108353

Keyword

easy,nonn,tabl

Author

Paul Barry, May 31 2005

Status

approved