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 A185410 A decomposition of the double factorials A001147. 4

%I

%S 1,1,0,1,2,0,1,10,4,0,1,36,60,8,0,1,116,516,296,16,0,1,358,3508,5168,

%T 1328,32,0,1,1086,21120,64240,42960,5664,64,0,1,3272,118632,660880,

%U 900560,320064,23488,128,0,1,9832,638968,6049744,14713840,10725184,2225728,95872,256,0

%N A decomposition of the double factorials A001147.

%C Row sums are A001147. Reversal of A185411.

%C From _Peter Bala_, Jul 24 2012: (Start)

%C This is the case k = 2 of the 1/k—Eulerian polynomials introduced by Savage and Viswanathan. They give a combinatorial interpretation of the triangle in terms of an ascent statistic on sets of inversion sequences and a geometric interpretation in terms of lecture hall polytopes.

%C Row reverse of A156919.

%C (End)

%C Triangle T(n,k), 0<=k<=n, given by (1, 0, 3, 0, 5, 0, 7, 0, 9, 0, ...) DELTA (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Feb 12 2013

%H G. C. Greubel, <a href="/A185410/b185410.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H S.-M. Ma, T. Mansour, <a href="http://arxiv.org/abs/1409.6525">The 1/k-Eulerian polynomials and k-Stirling permutations</a>, arXiv preprint arXiv:1409.6525 [math.CO], 2014.

%H C. D. Savage, G. Viswanathan, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i1p9">The 1/k-Eulerian polynomials</a>, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).

%F G.f.: 1/(1-x/(1-2xy/(1-3x/(1-4xy/(1-5x/(1-6xy/(1-7x/(1-8xy/(1- .... (continued fraction).

%F From _Peter Bala_, Jul 24 2012: (Start)

%F T(n,k) = sum {j=0..k}(-1)^(k-j)/4^j*C(n+1/2,k-j)*C(2*j,j)*(2*j+1)^n.

%F Recurrence equation: T(n+1,k) = (2*k+1)*T(n,k) + 2*(n-k+1)*T(n,k-1).

%F E.g.f.: sqrt(E(x,2*z)) = 1 + z + (1+2*x)*z^2/2! + (1+10*x+4*x^2)*z^3/3! + ..., where E(x,z) = (1-x)/(exp(z*(x-1)) - x) is the e.g.f. for the Eulerian numbers (version A173018). Cf. A156919.

%F Row polynomial R(n,x) = sum {k = 1..n} 2^(n-2*k)*C(2*k,k)*k!*Stirling2(n,k)*(x-1)^(n-k). R(n,4*x)/(1-4*x)^(n+1/2) = sum {k>=0} C(2*k,k)*(2*k+1)^n*x^k. The sequence of rational functions x*R(n,x)/(1-x)^(n+1) conjecturally occurs in the first column of (I - x*A112857)^(-1). (1+x)^(n-1)*R(n,x/(x+1)) gives the n-th row polynomial of A186695.

%F Row sums A001147. Alt. row sums A202038.

%F (End)

%F T(n,k) = 2^k*A102365(n,k). - _Philippe Deléham_, Feb 12 2013

%e Triangle begins:

%e 1,

%e 1, 0,

%e 1, 2, 0,

%e 1, 10, 4, 0,

%e 1, 36, 60, 8, 0,

%e 1, 116, 516, 296, 16, 0,

%e 1, 358, 3508, 5168, 1328, 32, 0,

%e 1, 1086, 21120, 64240, 42960, 5664, 64, 0,

%e 1, 3272, 118632, 660880, 900560, 320064, 23488, 128, 0,

%e 1, 9832, 638968, 6049744, 14713840, 10725184, 2225728, 95872, 256, 0,

%e ...

%e In the Savage-Viswanathan paper, the coefficients appear as

%e 1

%e 1 2

%e 1 10 4

%e 1 36 60 8

%e 1 116 516 296 16

%e 1 358 3508 5168 1328 32

%e 1 1086 21120 64240 42960 5664 64

%e ...

%t T[0, 0] := 1; T[n_, -1] := 0; T[n_, n_] := 0; T[n_, k_] := T[n, k] = (n - k)*T[n - 1, k - 1] + (2*k + 1)*T[n - 1, k]; Join[{1}, Table[If[k < 0, 0, If[k >= n, 0, 2^k*T[n, k]]], {n, 1, 5}, {k, 0, n}] // Flatten] (* _G. C. Greubel_, Jun 30 2017 *)

%Y Cf. A156919, A001147 (row sums), A112857, A173018, A186695, A202038 (alt. row sums).

%K nonn,easy,tabl

%O 0,5

%A _Paul Barry_, Jan 26 2011

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Last modified April 12 17:48 EDT 2021. Contains 342929 sequences. (Running on oeis4.)