A160563 Table of the number of (n,k)-Riordan complexes, read by rows.
1, 1, 1, 9, 10, 1, 225, 259, 35, 1, 11025, 12916, 1974, 84, 1, 893025, 1057221, 172810, 8778, 165, 1, 108056025, 128816766, 21967231, 1234948, 28743, 286, 1, 18261468225, 21878089479, 3841278805, 230673443, 6092515, 77077, 455, 1, 4108830350625, 4940831601000
Offset: 0
Examples
Triangle starts: [0] 1; [1] 1, 1; [2] 9, 10, 1; [3] 225, 259, 35, 1; [4] 11025, 12916, 1974, 84, 1; [5] 893025, 1057221, 172810, 8778, 165, 1; [6] 108056025, 128816766, 21967231, 1234948, 28743, 286, 1; . 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).
Links
- Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 2009.
- W. Zhang, Some identities involving the Euler and the central factorial numbers, The Fibonacci Quarterly, Vol. 36, Number 2, May 1998.
Programs
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Maple
t := proc(n,k) option remember ; expand(x*mul(x+n/2-i,i=1..n-1)) ; coeftayl(%,x=0,k) ; end: v := proc(n,k) option remember ; 4^(n-k)*t(2*n+1,2*k+1) ; end: 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 # Using a bivariate generating function (albeit generating signed terms): gf := (t + sqrt(1 + t^2))^x: ser := series(gf, t, 20): ct := n -> coeff(ser, t, n): T := (n, k) -> n!*coeff(ct(n), x, k): OddPart := (T, len) -> local n, k; seq(print(seq(T(n, k), k = 1..n, 2)), n = 1..2*len, 2): OddPart(T, 6); # Peter Luschny, Mar 03 2024
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Mathematica
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[n_, k_] := t[n, n - k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2017, after R. J. Mathar's comment *)
Formula
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).
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
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
Extensions
Extended by R. J. Mathar, May 20 2009
Comments