A209518 Triangle by rows, reversal of A104712.

1, 1, 3, 1, 4, 6, 1, 5, 10, 10, 1, 6, 15, 20, 15, 1, 7, 21, 35, 35, 21, 1, 8, 28, 56, 70, 56, 28, 1, 9, 36, 84, 126, 126, 84, 36, 1, 10, 45, 120, 210, 252, 210, 120, 45, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55

Offset: 0

Author

Gary W. Adamson, Mar 09 2012

Keywords

• nonn
• tabl

Comments

The offset is chosen as "0" to match the generalized or compositional Bernoulli numbers.

Following [Blandin and Diaz], we can generalize a subset of Bernoulli numbers to comply with the origin of the triangle (the Pascal matrix A007318 beheaded once: (A074909), twice: (this triangle), and so on...); and a corresponding Bernoulli sequence that equals the inverse of the triangle, extracting the left border. This procedure done with A074909 results in The Bernoulli numbers (A027641/A026642) starting (1, -1/2, 1/6,...). Done with this triangle we obtain A006568/A006569: (1, -1/3, 1/18, 1/90,...).

A generalized algebraic property of the subset of such triangles and compositional Bernoulli numbers is that the triangle M * [corresponding Bernoulli sequence considered as a vector, V] = [1, 0, 0, 0,...].

The infinite set of generalized Bernoulli number sequences thus generated from variants of Pascal's triangle begins: [(1, -1/2, 1/6,...); (1, -1/3, 1/18,...); (1, -1/4, 1/40,...); (1, -1/5, 1/75,...); where the third term denominators = A002411 (1, 6, 18, 40, 75,...) after the "1".

Row sums of the triangle = A000295 starting (1, 4, 11, 26, 57,...).

Examples

			First few rows of the triangle =
1;
1, 3;
1, 4, 6;
1, 5, 10, 10;
1, 6, 15, 20, 15;
1, 7, 21, 35, 35, 21;
1, 8, 28, 56, 70, 56, 28;
1, 9, 36, 84, 126, 126, 84, 36;
1, 10, 45, 120, 210, 252, 210, 120, 45;
1, 11, 55, 165, 330, 462, 462, 30, 165, 55;
...
		

Links

• Hector Blandin and Rafael Diaz, Compositional Bernoulli numbers, arXiv:0708.0809 [math.CO], 2007-2008, Page 7, 2nd table is identical to A006569/A006568.

Crossrefs

Cf. A000295, A104721, A074909, A027641, A027642, A006568, A006569, A007318, A002411.

Programs

Mathematica

Table[Binomial[n+2, k+2], {n, 0, 9}, {k, n , 0, -1}] // Flatten (* Jean-François Alcover, Aug 08 2018 *)

Formula

Doubly beheaded variant of Pascal's triangle in which two rightmost diagonals are deleted.

T(n,k)=T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-3*T(n-2,k-2)+T(n-3,k-2)+T(n-3,k-3), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014