

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.


7



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
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OFFSET

0,4


COMMENTS

T is an example of the group of matrices outlined in the table in A132382the associated matrix for aC(1,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) * (12*x)^(1/2). T(n,k) = Binomial(n,k)* s(nk) where s = A053871 with an e.g.f. of exp(x) * (12*x)^(1/2) which is the reciprocal of the e.g.f. of A055142. The row polynomials form an Appell sequence. Tom Copeland, Sep 10 2008


LINKS



FORMULA

T(n, k) = A053871(nk)*binomial(n, k).
E.g.f.: e^(xt) e^(t) (12t)^(1/2) = e^(p.(x)*t)(from my 2008 comment).
L = D = d/dx and R = x + d[log[e^(L)(12L)^(1/2)]]/dL = x  1 + 1/(12D) = x + 2D + (2D)^2 + (2D)^3 + ... are the lowering and raising operators, i.e., L p_n(x) = n * p_(n1)(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 (12D)^(1/2) e^(D) x^n = (12D)^(1/2) (x1)^n = p_n(x). For example, (12D)^(1/2)(x1)^2 = (1 + D + 3 D^2/2 + ...) (x1)^2 = (x1)^2 + 2(x1) + 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_(nk) * 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 (12t)^(1/2) associated with A055142. Cf. A231846 for a definition of umbral compositional inversion.
(End)


EXAMPLE

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;
...


MAPLE

g[0] := 1: g[1] := 0: for n from 2 to 20 do g[n] := (2*(n1))*(g[n1]+g[n2]) end do: T := proc (n, k) options operator, arrow; g[nk]*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


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 *)


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AUTHOR



STATUS

approved



