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A160563 Table of the number of (n,k)-Riordan complexes, read by rows.

%I A160563 #26 Jan 05 2025 19:51:39
%S A160563 1,1,1,9,10,1,225,259,35,1,11025,12916,1974,84,1,893025,1057221,
%T A160563 172810,8778,165,1,108056025,128816766,21967231,1234948,28743,286,1,
%U A160563 18261468225,21878089479,3841278805,230673443,6092515,77077,455,1,4108830350625,4940831601000
%N A160563 Table of the number of (n,k)-Riordan complexes, read by rows.
%C A160563 From Table 4, right-hand side, of Gelineau and Zeng.
%C A160563 Essentially a row-reversal of A008956. - _R. J. Mathar_, May 20 2009
%H A160563 Yoann Gelineau and Jiang Zeng, <a href="http://arxiv.org/abs/0905.2899"> Combinatorial Interpretations of the Jacobi-Stirling Numbers</a>, arXiv:0905.2899 [math.CO], May 2009.
%H A160563 W. Zhang, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/36-2/zhang.pdf">Some identities involving the Euler and the central factorial numbers</a>, The Fibonacci Quarterly, Vol. 36, Number 2, May 1998.
%F A160563 a(n,k) = |v(n,k)| where v(n,k) = v(n-1,k-1) - (2n-1)^2*v(n-1,k); eq (4.2).
%F A160563 Let F(x) = 1/cos(x). Then (2*n)!*(1/cos(x))^(2*n+1) = Sum_{k=0..n} T(n,k)*F^(2*k)(x), where F^(r) denotes the r-th derivative of F(x) (Zhang 1998). An example is given below. - _Peter Bala_, Feb 06 2012
%F A160563 Given a (0, 0)-based triangle U we call the triangle [U(n, k), k=1..n step 2, n=1..len step 2] the 'odd subtriangle' of U. This triangle is the odd subtriangle of U(n, k) = n! * [x^(n-k)] [t^n] (t + sqrt(1 + t^2))^x, albeit with signed terms. See A182867 for the even subtriangle. - _Peter Luschny_, Mar 03 2024
%e A160563 Triangle starts:
%e A160563   [0]         1;
%e A160563   [1]         1,          1;
%e A160563   [2]         9,         10,        1;
%e A160563   [3]       225,        259,       35,        1;
%e A160563   [4]     11025,      12916,     1974,       84,     1;
%e A160563   [5]    893025,    1057221,   172810,     8778,   165,    1;
%e A160563   [6] 108056025,  128816766, 21967231,  1234948, 28743,  286, 1;
%e A160563 .
%e A160563 For row 3: F(x) := 1/cos(x). Then 225*F(x) + 259*(d/dx)^2(F(x)) + 35*(d/dx)^4(F(x)) + (d/dx)^6(F(x)) = 720*(1/cos(x))^7, where F^(r) denotes the r-th derivative of F(x).
%p A160563 t := proc(n,k) option remember ; expand(x*mul(x+n/2-i,i=1..n-1)) ; coeftayl(%,x=0,k) ; end:
%p A160563 v := proc(n,k) option remember ; 4^(n-k)*t(2*n+1,2*k+1) ; end:
%p A160563 A160563 := proc(n,k) abs(v(n,k)) ; end: for n from 0 to 10 do for k from 0 to n do printf("%d,",A160563(n,k)) ; od: od: # _R. J. Mathar_, May 20 2009
%p A160563 # Using a bivariate generating function (albeit generating signed terms):
%p A160563 gf := (t + sqrt(1 + t^2))^x: ser := series(gf, t, 20):
%p A160563 ct := n -> coeff(ser, t, n): T := (n, k) -> n!*coeff(ct(n), x, k):
%p A160563 OddPart := (T, len) -> local n, k;
%p A160563 seq(print(seq(T(n, k), k = 1..n, 2)), n = 1..2*len, 2):
%p A160563 OddPart(T, 6);  # _Peter Luschny_, Mar 03 2024
%t A160563 t[_, 0] = 1; t[n_, n_] := t[n, n] = ((2*n - 1)!!)^2; t[n_, k_] := t[n, k] = (2*n - 1)^2*t[n - 1, k - 1] + t[n - 1, k];
%t A160563 T[n_, k_] := t[n, n - k];
%t A160563 Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 28 2017, after _R. J. Mathar_'s comment *)
%Y A160563 Cf. A001819, A008275, A008277, A160562, A091885, A182867.
%K A160563 nonn,tabl
%O A160563 0,4
%A A160563 _Jonathan Vos Post_, May 19 2009
%E A160563 Extended by _R. J. Mathar_, May 20 2009