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A123736
Triangle T(n,k) = Sum_{j=0..k/2} binomial(n-j-1,k-2*j), read by rows.
4
1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 3, 4, 3, 2, 1, 1, 0, 1, 4, 7, 7, 5, 3, 2, 1, 1, 0, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 0, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1, 0, 1, 7, 22, 41, 51, 46, 33, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1
OFFSET
1,8
COMMENTS
Row sums give: A000225
LINKS
Eric Weisstein's World of Mathematics, Steenrod Algebra
EXAMPLE
The triangle starts in row n=1 with columns 0 <= k < 2*n:
1, 0;
1, 1, 1, 0;
1, 2, 2, 1, 1, 0;
1, 3, 4, 3, 2, 1, 1, 0;
1, 4, 7, 7, 5, 3, 2, 1, 1, 0;
1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 0;
1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1, 0;
1, 7, 22, 41, 51, 46, 33, 21, 13, 8, 5, 3, 2, 1, 1, 0;
1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0;
MAPLE
seq(seq(sum(binomial(n-j-1, k-2*j), j=0..floor(k/2)), k=0..2*n-1), n=1..10); # G. C. Greubel, Sep 05 2019
MATHEMATICA
Table[Sum[Binomial[n-j-1, k-2*j], {j, 0, Floor[k/2]}], {n, 10}, {k, 0, 2*n-1}]//Flatten (* modified by G. C. Greubel, Sep 05 2019 *)
PROG
(PARI) T(n, k) = sum(j=0, k\2, binomial(n-j-1, k-2*j));
for(n=1, 10, for(k=0, 2*n-1, print1(T(n, k), ", "))) \\ G. C. Greubel, Sep 05 2019
(Magma) [&+[Binomial(n-j-1, k-2*j): j in [0..Floor(k/2)]]: k in [0..2*n-1], n in [1..10]]; // G. C. Greubel, Sep 05 2019
(Sage) [[sum(binomial(n-j-1, k-2*j) for j in (0..floor(k/2))) for k in (0..2*n-1)] for n in (1..10)] # G. C. Greubel, Sep 05 2019
(GAP) Flat(List([1..10], n-> List([0..2*n-1], k-> Sum([0..Int(k/2)], j-> Binomial(n-j-1, k-2*j) )))); # G. C. Greubel, Sep 05 2019
CROSSREFS
Cf. A136431 (antidiagonals), A027926 (row-reversed), A004006 (column m=3)
Sequence in context: A265890 A226920 A362759 * A358476 A185304 A081389
KEYWORD
nonn,easy,tabf
AUTHOR
Roger L. Bagula, Nov 14 2006
STATUS
approved