A161126 Triangle read by rows: T(n,k) is the number of involutions of {1,2,...,n} having k descents (n >= 1; 0 <= k < n).

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 12, 6, 1, 1, 9, 28, 28, 9, 1, 1, 12, 57, 92, 57, 12, 1, 1, 16, 105, 260, 260, 105, 16, 1, 1, 20, 179, 630, 960, 630, 179, 20, 1, 1, 25, 289, 1397, 3036, 3036, 1397, 289, 25, 1, 1, 30, 444, 2836, 8471, 12132, 8471, 2836, 444, 30, 1, 1, 36

Offset

1,5

Comments

Also number of ballot sequences of length n with k ascents; also number of standard Young tableaux with n cells such that there are k pairs of cells (v,v+1) with v+1 lying in a row below v. - Joerg Arndt, Feb 21 2014

See the Brualdi/Ma reference for the connection to A138177. - Joerg Arndt, Nov 02 2014

Links

Alois P. Heinz, Rows n = 1..141, flattened

Richard A. Brualdi, Shi-Mei Ma, Enumeration of involutions by descents and symmetric matrices, European Journal of Combinatorics, vol. 43, pp. 220-228, (January 2015).

J. Désarménien and D. Foata, Fonctions symétriques et séries hypergéometriques basiques multivariées, Bull. Soc. Math. France, 113, 1985, 3-22.

Samantha Dahlberg, Combinatorial Proofs of Identities Involving Symmetric Matrices, arXiv:1410.7356 [math.CO], 2014-2017.

I. M. Gessel and C. Reutenauer, Counting permutations with given cycle structure and descent set, J. Combin. Theory, Ser. A, 64, 1993, 189-215.

V. J. W. Guo and J. Zeng, The Eulerian distribution on involutions is indeed unimodal, J. Combin. Theory, Ser. A, 113, 2006, 1061-1071.

Formula

Sum_{k=1..n} T(n,k) = A000085(n) (row sums).

Sum_{k=0..n-1} k*T(n,k) = A161125(n).

Generating polynomial of row n is P(n,t) = (1-t)^(n+1) * Sum_{r>=0} t^r*Sum_{k=0..floor(n/2)} C(r(r+1)/2+k-1,k)*C(r+n-2k,n-2k) (see Eq. (2.5) in the Guo-Zeng paper; see first Maple program).

Recursive relation for n >= 3, k >= 0: n*T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) + [(k+1)^2 + n-2]*T(n-2,k) + [2k(n-k-1)-n+3]*T(n-2,k-1] + [(n-k)^2+n-2]*T(n-2,k-2) (see Eq. (2.4) in the Guo-Zeng paper; see 2nd Maple program).

Example

T(4,2)=4 because we have 1432, 2143, 4231, and 3214.
Triangle starts:
01: 1
02: 1,  1
03: 1,  2,   1
04: 1,  4,   4,    1
05: 1,  6,  12,    6,     1
06: 1,  9,  28,   28,     9,      1
07: 1, 12,  57,   92,    57,     12,      1
08: 1, 16, 105,  260,   260,    105,     16,      1
09: 1, 20, 179,  630,   960,    630,    179,     20,     1
10: 1, 25, 289, 1397,  3036,   3036,   1397,    289,    25,    1
11: 1, 30, 444, 2836,  8471,  12132,   8471,   2836,   444,   30,   1
12: 1, 36, 659, 5434, 21529,  42417,  42417,  21529,  5434,  659,  36,  1
13: 1, 42, 945, 9828, 50423, 132146, 181734, 132146, 50423, 9828, 945, 42, 1
...

Maple

P := proc (n) options operator, arrow: sort(simplify((1-t)^(n+1)*(sum(t^r*(sum(binomial((1/2)*r*(r+1)+k-1, k)*binomial(r+n-2*k, n-2*k), k = 0 .. floor((1/2)*n))), r = 0 .. infinity)))) end proc: for n to 12 do seq(coeff(P(n), t, j), j = 0 .. n-1) end do; # yields sequence in triangular form
T := proc(n, k) option remember; if k < 0 then 0 elif n <= k then 0 elif n = 1 and k = 0 then 1 elif n = 2 and k = 0 then 1 elif n = 2 and k = 1 then 1 else ((k+1)*T(n-1, k)+(n-k)*T(n-1, k-1)+((k+1)^2+n-2)*T(n-2, k)+(2*k*(n-k-1)-n+3)*T(n-2, k-1)+((n-k)^2+n-2)*T(n-2, k-2))/n end if end proc: for n to 12 do seq(T(n, k), k = 0 .. n-1) end do; # yields sequence in triangular form

Mathematica

P[n_, t_] := (1-t)^(n+1)*Sum[t^r*Binomial[n+r, n]*HypergeometricPFQ[{(1 - n)/2, -n/2, r(r+1)/2}, {(-n-r)/2, (1-n-r)/2}, 1], {r, 0, n}]; row[n_] := CoefficientList[P[n, t] + O[t]^n, t]; Table[row[n], {n, 1, 13}] // Flatten (* Jean-François Alcover, Dec 20 2016 *)

Crossrefs

Cf. A000085, A161125, A138177.

Sequence in context: A274310 A096806 A116672 * A128562 A034368 A361745

Adjacent sequences: A161123 A161124 A161125 * A161127 A161128 A161129

Keyword

nonn,tabl

Author

Emeric Deutsch, Jun 09 2009

Status

approved