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A052553
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Square array of binomial coefficients T(n,k) = binomial(n,k), n >= 0, k >= 0, read by upward antidiagonals.
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13
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1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 1, 0, 0, 1, 4, 3, 0, 0, 0, 1, 5, 6, 1, 0, 0, 0, 1, 6, 10, 4, 0, 0, 0, 0, 1, 7, 15, 10, 1, 0, 0, 0, 0, 1, 8, 21, 20, 5, 0, 0, 0, 0, 0, 1, 9, 28, 35, 15, 1, 0, 0, 0, 0, 0, 1, 10, 36, 56, 35, 6, 0, 0, 0, 0, 0, 0, 1, 11, 45, 84, 70, 21, 1, 0, 0, 0, 0, 0, 0, 1, 12, 55
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OFFSET
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0,8
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COMMENTS
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Another version of Pascal's triangle A007318.
As a triangle read by rows, it is (1,0,0,0,0,0,0,0,0,...) DELTA (0,1,-1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938 and it is the Riordan array (1/(1-x), x^2/(1-x)). The row sums of this triangle are F(n+1) = A000045(n+1). - Philippe Deléham, Dec 11 2011
As a triangle, binomial(n-k, k) is also the number of ways to add k pierced circles to a path graph P_n so that no two circles share a vertex (see Lemma 3.1 at page 5 in Owad and Tsvietkova). - Stefano Spezia, May 18 2022
For all n >= 0, k >= 0, the k-th homology group of the n-torus H_k(T^n) is the free abelian group of rank T(n,k) = binomial(n,k). See the Math Stack Exchange link below. - Jianing Song, Mar 13 2023
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LINKS
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FORMULA
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As a triangle: T(n,k) = A026729(n,n-k).
G.f. of the triangular version: 1/(1-x-x^2*y). - R. J. Mathar, Aug 11 2015
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EXAMPLE
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Array begins:
1, 0, 0, 0, 0, 0, ...
1, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 3, 3, 1, 0, 0, ...
1, 4, 6, 4, 1, 0, ...
1, 5, 10, 10, 5, 1, ...
As a triangle, this begins:
1;
1, 0;
1, 1, 0;
1, 2, 0, 0;
1, 3, 1, 0, 0;
1, 4, 3, 0, 0, 0;
1, 5, 6, 1, 0, 0, 0;
1, 6, 10, 4, 0, 0, 0, 0;
...
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MAPLE
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with(combinat): for s from 0 to 20 do for n from s to 0 by -1 do printf(`%d, `, binomial(n, s-n)) od:od: # James A. Sellers, Mar 17 2000
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MATHEMATICA
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PROG
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(Magma) /* As triangle */ [[Binomial(n-k, k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 08 2017
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CROSSREFS
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The official entry for Pascal's triangle is A007318. See also A026729 (the same array read by downward antidiagonals).
As triangle without zeroes: A011973.
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KEYWORD
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AUTHOR
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STATUS
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approved
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