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A124926
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Triangle read by rows: T(n,k) = binomial(n,k)*r(k), where r(k) are the Riordan numbers (r(k) = A005043(k); 0 <= k <= n).
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2
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1, 1, 0, 1, 0, 1, 1, 0, 3, 1, 1, 0, 6, 4, 3, 1, 0, 10, 10, 15, 6, 1, 0, 15, 20, 45, 36, 15, 1, 0, 21, 35, 105, 126, 105, 36, 1, 0, 28, 56, 210, 336, 420, 288, 91, 1, 0, 36, 84, 378, 756, 1260, 1296, 819, 232, 1, 0, 45, 120, 630, 1512, 3150, 4320, 4095, 2320, 603
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OFFSET
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0,9
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COMMENTS
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Row sums = Catalan numbers, A000108: (1, 1, 2, 5, 14, 42...); e.g., sum of row 4 terms = A000108(4) = 14 = (1 + 0 + 6 + 4 + 3). A005043 is the inverse binomial transform of the Catalan numbers.
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LINKS
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EXAMPLE
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First few rows of the triangle:
1;
1, 0;
1, 0, 1;
1, 0, 3, 1;
1, 0, 6, 4, 3;
1, 0, 10, 10, 15, 6;
1, 0, 15, 20, 45, 36, 15;
...
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MAPLE
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r:=n->(1/(n+1))*sum((-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i), i=0..n): T:=(n, k)->r(k)*binomial(n, k): for n from 0 to 12 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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MATHEMATICA
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T[n_, k_]:= T[n, k]= Binomial[n, k]*Sum[(-1)^j*Binomial[k+1, j]* Binomial[2*(k-j), k-j], {j, 0, k}]/(k+1); Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 19 2019 *)
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PROG
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(PARI) T(n, k) = b=binomial; b(n, k)*sum(j=0, k, (-1)^j*b(k+1, j)*b(2*(k-j), k-j))/(k+1); \\ G. C. Greubel, Nov 19 2019
(Magma) B:=Binomial; [B(n, k)*(&+[(-1)^j*B(k+1, j)*B(2*(k-j), k-j): j in [0..k]])/(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 19 2019
(Sage) b=binomial; [[b(n, k)*sum((-1)^j*b(k+1, j)*b(2*(k-j), k-j) for j in (0..k))/(k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 19 2019
(GAP) B:=Binomial;; Flat(List([0..12], n-> List([0..n], k-> B(n, k)* Sum([0..k], j-> (-1)^j*B(k+1, j)*B(2*(k-j), k-j))/(k+1) ))); # G. C. Greubel, Nov 19 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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