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%I #11 Sep 08 2022 08:45:19
%S 1,1,1,4,2,1,13,7,3,1,46,24,11,4,1,166,86,40,16,5,1,610,314,148,62,22,
%T 6,1,2269,1163,553,239,91,29,7,1,8518,4352,2083,920,367,128,37,8,1,
%U 32206,16414,7896,3544,1461,541,174,46,9,1,122464,62292,30086,13672,5776,2232
%N A tree-node counting triangle.
%C Columns include A026641,A014300,A014301. Inverse matrix is A109246. Row sums are A014300. Diagonal sums are A109245.
%H G. C. Greubel, <a href="/A109244/b109244.txt">Rows n=0..100 of triangle, flattened</a>
%F Number triangle T(n, k) = Sum_{i=0..n} (-1)^(n-i)*binomial(n+i-k, i-k).
%F Riordan array (1/(1-x*c(x)-2*x^2*c(x)^2), x*c(x)) where c(x)=g.f. of A000108.
%F The production matrix M (discarding the zeros) is:
%F 1, 1;
%F 3, 1, 1;
%F 3, 1, 1, 1;
%F 3, 1, 1, 1, 1;
%F ... such that the n-th row of the triangle is the top row of M^n. - Gary W. Adamson, Feb 16 2012
%e Rows begin
%e 1;
%e 1,1;
%e 4,2,1;
%e 13,7,3,1;
%e 46,24,11,4,1;
%e 166,86,40,16,5,1;
%t Table[Sum[(-1)^(n-j)*Binomial[n+j-k, j-k], {j,0,n}], {n,0,12}, {k,0,n}] //Flatten (* _G. C. Greubel_, Feb 19 2019 *)
%o (PARI) {T(n,k) = sum(j=0,n, (-1)^(n-j)*binomial(n+j-k, j-k))};
%o for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Feb 19 2019
%o (Magma) [[(&+[(-1)^(n-j)*Binomial(n+j-k, j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // _G. C. Greubel_, Feb 19 2019
%o (Sage) [[sum((-1)^(n-j)*binomial(n+j-k, j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Feb 19 2019
%o (GAP) Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j-> (-1)^(n-j)*Binomial(n+j-k, j-k) )))); # _G. C. Greubel_, Feb 19 2019
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, Jun 23 2005
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