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A058087
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Triangle read by rows, giving coefficients of the ménage hit polynomials ordered by descending powers. T(n, k) for 0 <= k <= n.
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11
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1, 2, -1, 2, 0, 0, 2, 3, 0, 1, 2, 8, 4, 8, 2, 2, 15, 20, 40, 30, 13, 2, 24, 60, 152, 210, 192, 80, 2, 35, 140, 469, 994, 1477, 1344, 579, 2, 48, 280, 1232, 3660, 7888, 11672, 10800, 4738, 2, 63, 504, 2856, 11268, 32958, 70152, 104256, 97434, 43387
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OFFSET
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0,2
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COMMENTS
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Riordan's book (page 197) notes that an alternative convention is to put 2 in the first row of the triangle. - William P. Orrick, Aug 09 2020
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REFERENCES
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I. Kaplansky and J. Riordan, The probleme des menages, Scripta Mathematica, 1946, 12 (2), 113-124.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 198.
Tolman, L. Kirk, "Extensions of derangements", Proceedings of the West Coast Conference on Combinatorics, Graph Theory and Computing, Humboldt State University, Arcata, California, September 5-7, 1979. Vol. 26. Utilitas Mathematica Pub., 1980. See Table I. - N. J. A. Sloane, Jul 06 2014
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LINKS
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FORMULA
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G.f.: (1-x*(y-1))*Sum_{n>=0} ( n!*(x*y)^n/(1+x*(y-1))^(2*n+1) ). - Vladeta Jovovic, Dec 14 2009
Row n of the triangle lists the coefficients of the polynomial U_n(t) = Sum_{k=0..n} ((2*n/(2*n-k))*binomial(2*n-k,k)*(n-k)!*(t-1)^k, with higher order terms first (Kaplansky and Riordan). - William P. Orrick, Aug 09 2020
T(n, k) = Sum_{j=0..k} (-1)^j*(2*n*(k-j)!/(n+k-j))*binomial(n-k+j, n-k)*binomial(n+k-j, n-k+j), with T(0, k) = 1. - G. C. Greubel, May 15 2021 [Corrected by Sean A. Irvine, Jul 23 2022]
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EXAMPLE
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The triangle begins:
1;
2, -1;
2, 0, 0;
2, 3, 0, 1;
2, 8, 4, 8, 2;
2, 15, 20, 40, 30, 13;
2, 24, 60, 152, 210, 192, 80;
2, 35, 140, 469, 994, 1477, 1344, 579;
2, 48, 280, 1232, 3660, 7888, 11672, 10800, 4738;
2, 63, 504, 2856, 11268, 32958, 70152, 104256, 97434, 43387;
The polynomials start:
[0] 1;
[1] 2*x - 1;
[2] 2*x^2;
[3] 2*x^3 + 3*x^2 + 1;
[4] 2*x^4 + 8*x^3 + 4*x^2 + 8*x + 2;
[5] 2*x^5 + 15*x^4 + 20*x^3 + 40*x^2 + 30*x + 13.
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MAPLE
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U := proc(n) if n = 0 then return 1 fi;
add((2*n/(2*n-k))*binomial(2*n-k, k)*(n-k)!*(x-1)^k, k=0..n) end:
W := proc(r, s) coeff(U(r), x, s ) end:
T := (n, k) -> W(n, n-k): seq(seq(T(n, k), k=0..n), n=0..9);
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MATHEMATICA
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u[n_] := Sum[ 2*n/(2*n-k)*Binomial[2*n-k, k]*(n-k)!*(x-1)^k, {k, 0, n}]; w[r_, s_] := Coefficient[u[r], x, s]; a[n_, k_] := w[n, n-k]; a[0, 0]=1; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 10 2012, translated from Maple *)
T[n_, k_]:= If[n==0, 1, Sum[(-1)^j*(2*n*(k-j)!/(n+k-j))*Binomial[j+n-k, n - k]*Binomial[n+k-j, n-k+j], {j, 0, k}]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 15 2021 *)
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PROG
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(SageMath)
a = [[1]]
for n in range(1, 10):
g = expand(
sum((x - 1)^ k * (2*n) * binomial(2*n-k, k) * factorial(n-k) / (2*n-k)
for k in range(0, n + 1)
)
)
coeffs = g.coefficients(sparse=False)
coeffs.reverse()
(Sage)
def A058087(n, k): return 1 if (n==0) else sum( (-1)^j*(2*n*factorial(k-j)/(n+k-j))*binomial(j+n-k, n-k)*binomial(n+k-j, n-k+j) for j in (0..k) )
(PARI) U(n, t)=sum(k=0, n, ((2*n/(2*n-k))*binomial(2*n-k, k)*(n-k)!*(t-1)^k));
print1(1, ", "); for(n=1, 9, forstep(k=n, 0, -1, print1(polcoef(U(n, 'x), k), ", "))) \\ Hugo Pfoertner, Aug 30 2020
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CROSSREFS
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Essentially a mirror image of A094314.
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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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