A161123 - OEIS

A161123

Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k inversions (0 <= k <= n(2n-1)).

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 3, 0, 2, 0, 1, 0, 0, 0, 0, 1, 0, 3, 0, 6, 0, 9, 0, 12, 0, 14, 0, 15, 0, 14, 0, 12, 0, 9, 0, 6, 0, 3, 0, 1, 0, 0, 0, 0, 0, 1, 0, 4, 0, 10, 0, 19, 0, 31, 0, 45, 0, 60, 0, 74, 0, 86, 0, 94, 0, 97, 0, 94, 0, 86, 0, 74, 0, 60, 0, 45, 0, 31, 0, 19

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OFFSET

0,16

COMMENTS

Sum of entries in row n is (2n-1)!! = A001147(n).

Row n has 1 + 2n(n-1) entries.

Sum_{k>=0} k*T(n,k) = (2n-1)!!*n^2 = A161124(n).

A128080 is the same triangle with the 0's deleted.

LINKS

Table of n, a(n) for n=0..94.

FORMULA

Generating polynomial of row n is P_n(q) = (q/(1-q^2))^n*Product_{j=1..n}(1-q^(4j-2)).

EXAMPLE

T(3,11)=3 because we have 465132, 546213, and 632541.

Triangle starts:

0, 1;

0, 0, 1, 0, 1, 0, 1;

0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 3, 0, 2, 0, 1;

MAPLE

f := proc (n) options operator, arrow: q^n*(product(1-q^(4*j-2), j = 1 .. n))/(1-q^2)^n end proc: for n from 0 to 4 do P[n] := sort(expand(simplify(f(n)))) end do: for n from 0 to 4 do seq(coeff(P[n], q, j), j = 0 .. n*(2*n-1)) end do; # yields sequence in triangular form

MATHEMATICA

P[n_] := P[n] = q^n*Product[1 - q^(4j - 2), {j, 1, n}]/(1 - q^2)^n // Expand // Simplify;

T[n_, k_] := Coefficient[P[n], q, k];

Table[T[n, k], {n, 0, 5}, {k, 0, n (2n - 1)}] // Flatten (* Jean-François Alcover, Aug 23 2024 *)

CROSSREFS

Cf. A001147, A128080, A161124.

Sequence in context: A189230 A243982 A214000 * A035442 A213177 A265017

Adjacent sequences: A161120 A161121 A161122 * A161124 A161125 A161126

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, Jun 05 2009

STATUS

approved