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Triangle read by rows: T(n,k) is the number of involutions of {1,2,...,n} having k descents (n >= 1; 0 <= k < n).
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%I #41 Jan 22 2020 20:09:41

%S 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,

%T 105,260,260,105,16,1,1,20,179,630,960,630,179,20,1,1,25,289,1397,

%U 3036,3036,1397,289,25,1,1,30,444,2836,8471,12132,8471,2836,444,30,1,1,36

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

%C 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

%C See the Brualdi/Ma reference for the connection to A138177. - _Joerg Arndt_, Nov 02 2014

%H Alois P. Heinz, <a href="/A161126/b161126.txt">Rows n = 1..141, flattened</a>

%H Richard A. Brualdi, Shi-Mei Ma, <a href="https://doi.org/10.1016/j.ejc.2014.08.026">Enumeration of involutions by descents and symmetric matrices</a>, European Journal of Combinatorics, vol. 43, pp. 220-228, (January 2015).

%H J. Désarménien and D. Foata, <a href="http://www.numdam.org/item?id=BSMF_1985__113__3_0">Fonctions symétriques et séries hypergéometriques basiques multivariées</a>, Bull. Soc. Math. France, 113, 1985, 3-22.

%H Samantha Dahlberg, <a href="http://arxiv.org/abs/1410.7356">Combinatorial Proofs of Identities Involving Symmetric Matrices</a>, arXiv:1410.7356 [math.CO], 2014-2017.

%H I. M. Gessel and C. Reutenauer, <a href="http://dx.doi.org/10.1016/0097-3165(93)90095-P">Counting permutations with given cycle structure and descent set</a>, J. Combin. Theory, Ser. A, 64, 1993, 189-215.

%H V. J. W. Guo and J. Zeng, <a href="http://dx.doi.org/10.1016/j.jcta.2005.10.002">The Eulerian distribution on involutions is indeed unimodal</a>, J. Combin. Theory, Ser. A, 113, 2006, 1061-1071.

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

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

%F 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).

%F 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).

%e T(4,2)=4 because we have 1432, 2143, 4231, and 3214.

%e Triangle starts:

%e 01: 1

%e 02: 1, 1

%e 03: 1, 2, 1

%e 04: 1, 4, 4, 1

%e 05: 1, 6, 12, 6, 1

%e 06: 1, 9, 28, 28, 9, 1

%e 07: 1, 12, 57, 92, 57, 12, 1

%e 08: 1, 16, 105, 260, 260, 105, 16, 1

%e 09: 1, 20, 179, 630, 960, 630, 179, 20, 1

%e 10: 1, 25, 289, 1397, 3036, 3036, 1397, 289, 25, 1

%e 11: 1, 30, 444, 2836, 8471, 12132, 8471, 2836, 444, 30, 1

%e 12: 1, 36, 659, 5434, 21529, 42417, 42417, 21529, 5434, 659, 36, 1

%e 13: 1, 42, 945, 9828, 50423, 132146, 181734, 132146, 50423, 9828, 945, 42, 1

%e ...

%p 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

%p 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

%t 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 *)

%Y Cf. A000085, A161125, A138177.

%K nonn,tabl

%O 1,5

%A _Emeric Deutsch_, Jun 09 2009