A055140 Triangle read by rows: T(n, k) = number of matchings of 2n people with partners (of either sex) such that exactly k couples are left together.
1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 60, 32, 12, 0, 1, 544, 300, 80, 20, 0, 1, 6040, 3264, 900, 160, 30, 0, 1, 79008, 42280, 11424, 2100, 280, 42, 0, 1, 1190672, 632064, 169120, 30464, 4200, 448, 56, 0, 1, 20314880, 10716048, 2844288, 507360, 68544, 7560, 672, 72, 0, 1
Offset: 0
Examples
Triangle T(n,k) starts: 1; 0, 1; 2, 0, 1; 8, 6, 0, 1; 60, 32, 12, 0, 1; 544, 300, 80, 20, 0, 1; 6040, 3264, 900, 160, 30, 0, 1; ...
Links
- Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
- Donovan Young, Generating Functions for Domino Matchings in the 2 * k Game of Memory, arXiv:1905.13165 [math.CO], 2019. Also in J. Int. Seq., Vol. 22 (2019), Article 19.8.7.
Crossrefs
Programs
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Maple
g[0] := 1: g[1] := 0: for n from 2 to 20 do g[n] := (2*(n-1))*(g[n-1]+g[n-2]) end do: T := proc (n, k) options operator, arrow; g[n-k]*binomial(n, k) end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jan 24 2009
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Mathematica
Table[(-1)^# HypergeometricPFQ[{1/2, -#}, {}, 2] Binomial[n, k] &[n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 10 2019, after Eric W. Weisstein at A053871 *)
Formula
T(n, k) = A053871(n-k)*binomial(n, k).
From Tom Copeland, Oct 12 2016: (Start)
E.g.f.: e^(xt) e^(-t) (1-2t)^(-1/2) = e^(p.(x)*t)(from my 2008 comment).
Row sums are A001147.
L = D = d/dx and R = x + d[log[e^(L)(1-2L)^(-1/2)]]/dL = x - 1 + 1/(1-2D) = x + 2D + (2D)^2 + (2D)^3 + ... are the lowering and raising operators, i.e., L p_n(x) = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x); e.g., L p_2(x) = D (2 + x^2) = 2 x = 2 p_1(x) and R P_2(x) = (x + 2D + 4D^2 + ...) (2 + x^2) = 2x + x^3 + 4x + 8 = 8 + 6x + x^3 = p_3(x).
Another generator is (1-2D)^(-1/2) e^(-D) x^n = (1-2D)^(-1/2) (x-1)^n = p_n(x). For example, (1-2D)^(-1/2)(x-1)^2 = (1 + D + 3 D^2/2 + ...) (x-1)^2 = (x-1)^2 + 2(x-1) + 3 = 2 + x^2 = p_2(x).
Umbral binomial convolution gives p_n(x) = (a. + x)^n = sum_{k = 0,..,n} C(n,k) a_(n-k) * x^k with (a.)^k = a_k = A053871(k).
The Appell sequence of umbral compositional inverses has the e.g.f. e^(xt) e^t (1-2t)^(1/2) associated with A055142. Cf. A231846 for a definition of umbral compositional inversion.
(End)
Comments