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A080928
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Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).
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9
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1, 1, 0, 2, 2, 1, 4, 6, 3, 0, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 0, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 0, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10
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OFFSET
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0,4
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COMMENTS
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Gives the general solution to a(n) = 2*a(n-1) + k(k+2)*a(n-2), a(0) = a(1) = 1. The value k=1 gives the row sums of the triangle, or 1,1,5,13,... This is A046717, the solution to a(n) = 2*a(n-1) + 3*a(n-2), a(0)=a(1)=1.
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 156.
J-L. Kim, Relation between weight distribution and combinatorial identities, Bulletin of the Institute of Combinatorics and its Applications, Canada, 31, 2001, pp. 69-79.
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LINKS
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FORMULA
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T(n, n) = (n+1) mod 2, T(n, k) = C(n, k)*2^(n-k-1).
T(n, 0) = A011782(n), T(n, k)=0, k>n, T(2n, 2n)=1, T(2n-1, 2n-1)=0, T(n+1, n)=n+1. Otherwise T(n, k) = T(n-1, k-1) + 2T(n-1, k). Rows are the coefficients of the polynomials in the expansion of (1-x)/((1+kx)(1-(k+2)x). The main diagonal is 1, 0, 1, 0, 1, 0, ... with g.f. 1/(1-x^2). Subsequent subdiagonals are given by A011782(k)*C(n+k, k) with g.f. A011782(k)/(1-x)^k.
T(n, k) = Sum_{j=0..n} C(n, j)*C(j, k)*(1+(-1)^j)/2; T(n, k) = 2^(n-k-1)*(C(n, k) + (-1)^n*C(0, n-k)). - Paul Barry, Nov 08 2005
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EXAMPLE
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Triangle begins:
1;
1, 0;
2, 2, 1;
4, 6, 3, 0;
8, 16, 12, 4, 1;
16, 40, 40, 20, 5, 0;
32, 96, 120, 80, 30, 6, 1;
64, 224, 336, 280, 140, 42, 7, 0;
128, 512, 896, 896, 560, 224, 56, 8, 1;
256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0; etc.
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MATHEMATICA
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Table[Sum[Binomial[n, 2 i] Binomial[2 i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 11 2018 *)
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CROSSREFS
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Apart from k=n, T(n, k) equals (1/2)*A038207(n, k).
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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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