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Number triangle T(n,k) = Sum_{j=0..n-k} C(k,2j)*C(n-k,2j).
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%I #21 Mar 30 2023 16:08:30

%S 1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,4,4,1,1,1,1,7,10,7,1,1,1,1,11,19,

%T 19,11,1,1,1,1,16,31,38,31,16,1,1,1,1,22,46,66,66,46,22,1,1,1,1,29,64,

%U 106,126,106,64,29,1,1,1,1,37,85,162,226,226,162,85,37,1,1

%N Number triangle T(n,k) = Sum_{j=0..n-k} C(k,2j)*C(n-k,2j).

%C Third column is essentially A000124. Fourth column is essentially A005448. Fifth column is A119327. Product of Pascal's triangle A007318 and A119328. Row sums are A038504. T(n,k) = T(n,n-k).

%D Lukas Spiegelhofer and Jeffrey Shallit, Continuants, Run Lengths, and Barry's Modified Pascal Triangle, Volume 26(1) 2019, of The Electronic Journal of Combinatorics, #P1.31.

%H Seiichi Manyama, <a href="/A119326/b119326.txt">Rows n = 0..139, flattened</a>

%H Jeffrey Shallit, Lukas Spiegelhofer, <a href="https://arxiv.org/abs/1710.06203">Continuants, run lengths, and Barry's modified Pascal triangle</a>, arXiv:1710.06203 [math.CO], 2017.

%F Column k has g.f.: (x^k/(1-x))* Sum{j=0..k} C(k,2j)*(x/(1-x))^(2j).

%F T(2n,n) = A119358(n). - _Alois P. Heinz_, Aug 31 2018

%e Triangle begins:

%e 1;

%e 1, 1;

%e 1, 1, 1;

%e 1, 1, 1, 1;

%e 1, 1, 2, 1, 1;

%e 1, 1, 4, 4, 1, 1;

%e 1, 1, 7, 10, 7, 1, 1;

%e 1, 1, 11, 19, 19, 11, 1, 1;

%e ...

%Y Cf. A119358.

%K easy,nonn,tabl

%O 0,13

%A _Paul Barry_, May 14 2006