A363418 Square array read by ascending antidiagonals: T(n,k) = [x^(n*k)] ((1 + x)/(1 - x))^k for n, k >= 1.
2, 2, 8, 2, 16, 38, 2, 24, 146, 192, 2, 32, 326, 1408, 1002, 2, 40, 578, 4672, 14002, 5336, 2, 48, 902, 11008, 69002, 142000, 28814, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146
Offset: 1
Examples
Square array begins n\k | 1 2 3 4 5 6 7 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 | 2 8 38 192 1002 5336 28814 ... (A002003) 2 | 2 16 146 1408 14002 142000 1459810 ... (A103885) 3 | 2 24 326 4672 69002 1038984 15856206 ... (A333715) 4 | 2 32 578 11008 216002 4320608 87588482 ... 5 | 2 40 902 21440 525002 13104184 331482062 ... 6 | 2 48 1298 36992 1086002 32497680 985524066 ... 7 | 2 56 1766 58688 2009002 70097384 2478629134 ... 8 | 2 64 2306 87552 3424002 136485568 5513464322 ...
References
- R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 antidiagonals)
Crossrefs
Programs
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Maple
# display as a square array T := (n,k) -> add( binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j) , j = 0..k): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do; #alternative program seq(print(seq(simplify(2*k*hypergeom([1 - k, 1 - n*k], [2], 2)), k = 1..10)), n = 1..10); # display as a sequence seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);
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PARI
T(n,k) = sum(j=0, k, binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j)) \\ Andrew Howroyd, Jan 05 2024
Formula
T(n,k) = Sum_{j = 0..k} binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j).
T(n,k) = 1/n * [x^k] ((1 + x)/(1 - x))*(n*k).
T(n,k) = (1/n)*Sum_{j = 0..k} binomial(n*k, j)*binomial((n + 1)*k - j - 1, k - j).
T(2*n,k) = [x^(n*k)] Chebyshev_T(k,(1 + x)/(1 - x)), where Chebyshev_T(n,x) denotes the n-th Chebyshev polynomial of the first kind. See A053120.
T(n,k) = Sum_{j = 1..k} (2^j)*binomial(k, j)*binomial(n*k - 1, n*k - j).
T(n,k) = (2*k) * hypergeom([1 - k, 1 - n*k], [2], 2).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)/(1 + x))^n.
E(m,x) appears to be the g.f. of the (m + 1)-Schroeder numbers. See A027307 (m = 2) and the cross references there.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = (1 + x)/(1 - x), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
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