

A209518


Triangle by rows, reversal of A104712.


1



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
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OFFSET

0,3


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,...).


LINKS



FORMULA

Doubly beheaded variant of Pascal's triangle in which two rightmost diagonals are deleted.
T(n,k)=T(n1,k)+3*T(n1,k1)2*T(n2,k1)3*T(n2,k2)+T(n3,k2)+T(n3,k3), T(0,0)=1, T(n,k)=0 if k<0 or if k>n.  Philippe Deléham, Jan 11 2014


EXAMPLE

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;
...


MATHEMATICA



CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



