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A145882 Triangle read by rows: T(n,k) is the number of even permutations of {1,2,...,n} having k descents (n >= 1, k >= 0).

%I A145882 #48 Mar 21 2025 18:09:24
%S A145882 1,1,1,2,1,5,5,1,1,14,30,14,1,1,29,147,155,28,1,64,586,1208,605,56,1,
%T A145882 127,2133,7819,7819,2133,127,1,1,262,7288,44074,78190,44074,7288,262,
%U A145882 1,1,517,23893,227569,655315,655039,227623,23947,496,1,1044,76332,1101420
%N A145882 Triangle read by rows: T(n,k) is the number of even permutations of {1,2,...,n} having k descents (n >= 1, k >= 0).
%C A145882 Number of entries in row n is 1+floor(binomial(n,2)/2)-floor(binomial(n-2,2)/2).
%C A145882 Sum of entries in row n is A001710(n) for n>=2.
%H A145882 Alois P. Heinz, <a href="/A145882/b145882.txt">Rows n = 1..142, flattened</a>
%H A145882 J. Shareshian and M. L. Wachs, <a href="https://doi.org/10.1090/S1079-6762-07-00172-2">q-Eulerian polynomials: excedance number and major index</a>, Electronic Research Announcements of the Amer. Math. Soc., 13 (2007), 33-45.
%H A145882 R. P. Stanley, <a href="http://dx.doi.org/10.1016/0097-3165(76)90028-5">Binomial posets, Möbius inversion and permutation enumeration</a>, J. Combinat. Theory, A 20 (1976), 336-356.
%H A145882 S. Tanimoto, <a href="http://www.emis.de/journals/INTEGERS/papers/g31/g31.Abstract.html">A study of Eulerian numbers for permutations in the alternating group</a>, Integers, Electronic J. of Combinatorial Number Theory, 6 (2006), #A31.
%F A145882 In the Shareshian and Wachs reference (p. 35) a q-analog of the exponential g.f. of the Eulerian polynomials is given for the joint distribution of (inv, des) (see also the Stanley reference). The first Maple program given below makes use of this function by considering its even part.
%F A145882 T(n,k) = (euler(n,k) + Sum_{j=max(0, k+1-ceiling(n/2))..min(floor(n/2), k)} binomial(j-1-floor(n/2), j) * euler(ceiling(n/2), k-j)) / 2, where euler(n,k) is the Eulerian number A173018 (not A008292, which has different indexing). - _Robert A. Russell_, Nov 15 2018
%e A145882 T(4,2) = 5 because we have 4132, 2143, 4213, 2431 and 3241.
%e A145882 Triangle begins with T(1,0):
%e A145882   1
%e A145882   1
%e A145882   1    2
%e A145882   1    5      5       1
%e A145882   1   14     30      14       1
%e A145882   1   29    147     155      28
%e A145882   1   64    586    1208     605      56
%e A145882   1  127   2133    7819    7819    2133     127       1
%e A145882   1  262   7288   44074   78190   44074    7288     262     1
%e A145882   1  517  23893  227569  655315  655039  227623   23947   496
%e A145882   1 1044  76332 1101420 4869558 7862124 4868556 1102068 76305 992
%p A145882 for n to 11 do qbr := proc (m) options operator, arrow; sum(q^i, i = 0 .. m-1) end proc; qfac := proc (m) options operator, arrow; product(qbr(j), j = 1 .. m) end proc; Exp := proc (z) options operator, arrow; sum(q^binomial(m, 2)*z^m/qfac(m), m = 0 .. 19) end proc; g := (1-t)/(Exp(z*(t-1))-t); gser := simplify(series(g, z = 0, 17)); a[n] := simplify(qfac(n)*coeff(gser, z, n)); b[n] := (a[n]+subs(q = -q, a[n]))*1/2; P[n] := sort(subs(q = 1, b[n])) end do; for n to 11 do seq(coeff(P[n], t, j), j = 0 .. floor((1/2)*binomial(n, 2)) -floor((1/2)*binomial(n-2, 2))) end do; # yields sequence in triangular form
%p A145882 # second Maple program:
%p A145882 b:= proc(u, o, t) option remember; `if`(u+o=0, t, expand(
%p A145882        add(b(u+j-1, o-j, irem(t+j-1+u, 2)), j=1..o)+
%p A145882        add(b(u-j, o+j-1, irem(t+u-j, 2))*x, j=1..u)))
%p A145882     end:
%p A145882 T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))
%p A145882         (add(b(j-1, n-j, irem(j, 2)), j=1..n)):
%p A145882 seq(T(n), n=1..12);  # _Alois P. Heinz_, Nov 19 2013
%t A145882 b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, t, Expand[Sum[b[u+j-1, o-j, Mod[t+j-1+u, 2]], {j, 1, o}] + Sum[b[u-j, o+j-1, Mod[t+u-j, 2]]*x, {j, 1, u}]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]] [Sum[b[j-1, n-j, Mod[j, 2]], {j, 1, n}]]; Table[T[n], {n, 1, 12}] // Flatten (* _Jean-François Alcover_, May 26 2015, after _Alois P. Heinz_ *)
%t A145882 Needs["Combinatorica`"];
%t A145882 Table[(Eulerian[n, k] + Sum[Binomial[j-1-Floor[n/2], j] Eulerian[Ceiling[n/2], k-j], {j, Max[0, k-Ceiling[n/2]], Min[Floor[n/2], k]}])/2, {n, 25}, {k, 0, n-1}] // Flatten // DeleteCases[0] (* _Robert A. Russell_, Nov 14 2018 *)
%Y A145882 Cf. A001710, A145883.
%Y A145882 Cf. A128612 (similar with rows reversed).
%K A145882 nonn,tabf
%O A145882 1,4
%A A145882 _Emeric Deutsch_, Nov 11 2008