A039717 Row sums of convolution triangle A030523.

1, 4, 15, 55, 200, 725, 2625, 9500, 34375, 124375, 450000, 1628125, 5890625, 21312500, 77109375, 278984375, 1009375000, 3651953125, 13212890625, 47804687500, 172958984375, 625771484375, 2264062500000, 8191455078125

Offset: 1

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.

With offset 0 = INVERT transform of A001792: (1, 3, 8, 20, 48, 112, ...). - Gary W. Adamson, Oct 26 2010

From Tom Copeland, Nov 09 2014: (Start)

The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results).

Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).

O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).

Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)

p-INVERT of (2^n), where p(s) = 1 - s - s^2; see A289780. - Clark Kimberling, Aug 10 2017

Links

• Michel Marcus, Table of n, a(n) for n = 1..1000
• Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
• Index entries for linear recurrences with constant coefficients, signature (5,-5).

Crossrefs

Appears in A109106.

Cf. A000045, A000108, A001792, A005043, A091867, A026378, A030528.

Programs

Mathematica

CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)

PARI

Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ Altug Alkan, Nov 20 2015

Formula

G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).

From Paul Barry, Apr 16 2004: (Start)

Binomial transform of Fibonacci(2n+2).

a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)

a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2*n).

a(n) = 5*a(n-1) - 5*a(n-2).

a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - Paul Barry, Jun 22 2004

From Johannes W. Meijer, Jul 01 2010: (Start)

Limit_{k->oo} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.

Limit_{n->oo} A020876(n)/A093131(n) = sqrt(5).

(End)

From Benito van der Zander, Nov 19 2015: (Start)

Limit_{k->oo} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.

a(n) = a(n-1) * 3 + A081567(n-2) (not proved).

(End)

E.g.f.: exp(x*5/2) * (cosh(x*sqrt(5)/2) + (3/sqrt(5))*sinh(x*sqrt(5)/2)). - Fabian Pereyra, Oct 29 2024