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A109244
A tree-node counting triangle.
3
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, 6, 1, 2269, 1163, 553, 239, 91, 29, 7, 1, 8518, 4352, 2083, 920, 367, 128, 37, 8, 1, 32206, 16414, 7896, 3544, 1461, 541, 174, 46, 9, 1, 122464, 62292, 30086, 13672, 5776, 2232
OFFSET
0,4
COMMENTS
Columns include A026641,A014300,A014301. Inverse matrix is A109246. Row sums are A014300. Diagonal sums are A109245.
FORMULA
Number triangle T(n, k) = Sum_{i=0..n} (-1)^(n-i)*binomial(n+i-k, i-k).
Riordan array (1/(1-x*c(x)-2*x^2*c(x)^2), x*c(x)) where c(x)=g.f. of A000108.
The production matrix M (discarding the zeros) is:
1, 1;
3, 1, 1;
3, 1, 1, 1;
3, 1, 1, 1, 1;
... such that the n-th row of the triangle is the top row of M^n. - Gary W. Adamson, Feb 16 2012
EXAMPLE
Rows begin
1;
1,1;
4,2,1;
13,7,3,1;
46,24,11,4,1;
166,86,40,16,5,1;
MATHEMATICA
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 *)
PROG
(PARI) {T(n, k) = sum(j=0, n, (-1)^(n-j)*binomial(n+j-k, j-k))};
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Feb 19 2019
(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
(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
(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
CROSSREFS
Sequence in context: A152818 A302235 A242861 * A171650 A225476 A143777
KEYWORD
easy,nonn,tabl
AUTHOR
Paul Barry, Jun 23 2005
STATUS
approved