A155587 Expansion of (1 + x*c(x))/(1 - x), where c(x) is the g.f. of A000108.

1, 2, 3, 5, 10, 24, 66, 198, 627, 2057, 6919, 23715, 82501, 290513, 1033413, 3707853, 13402698, 48760368, 178405158, 656043858, 2423307048, 8987427468, 33453694488, 124936258128, 467995871778, 1757900019102, 6619846420554

Offset: 0

Author

Paul Barry, Jan 24 2009

Keywords

• easy
• nonn

Comments

Row sums of A155586.

Hankel transform is A057079(n+2).

From Petros Hadjicostas, Aug 03 2020: (Start)

To prove R. J. Mathar's conjecture, note that the o.g.f. of the sequence implies (Sum_{n >= 0} a(n)*x^n)*(1 - x) = 1 + x*c(x); i.e., a(0) + Sum_{n >= 1} (a(n) - a(n-1))*x^n = 1 + Sum_{n >= 1} C(n-1)*x^n, where C(n) = A000108(n) (Catalan numbers).

Thus, C(n-1) = a(n) - a(n-1) (for n >= 1), and hence C(n) = a(n+1) - a(n). Since 2*(2*n - 1)*C(n-1) = (n + 1)*C(n), we get (n + 1)*a(n+1) + (-5*n + 1)*a(n) + 2*(2*n - 1)*a(n-1) = 0. The last equation implies R. J. Mathar's conjecture. (End)

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Gerard Cohen and Jean-Pierre Flori, On a generalized combinatorial conjecture involving addition mod 2^k - 1, IACR, Report 2011/400.
• Jean-Pierre Flori, Fonctions booléennes, courbes algébriques et multiplication complexe, Thesis, ParisTech, Feb 03 2012.
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Crossrefs

Cf. A000108, A014138, A057079, A155586.

Partial sums of A120588.

Programs

Haskell

a155587 n = a155587_list !! n
a155587_list = scanl (+) 1 a000108_list  -- Reinhard Zumkeller, Mar 01 2013

Maple

CatalanNumber := n -> binomial(2*n, n)/(n+1):
a := n -> ((3 - I*sqrt(3)))/2 - CatalanNumber(n)*hypergeom([1, n+1/2], [n+2], 4):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Aug 04 2020

Formula

a(n) = 1 + Sum_{k=0..n-1} A000108(k).

Conjecture: n*a(n) + (6-5*n)*a(n-1) + 2*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 15 2011

a(n) = A014138(n-1) + 2 for n > 0. - Reinhard Zumkeller, Mar 01 2013 [Corrected by Petros Hadjicostas, Aug 03 2020]

a(n+1) - a(n) = A000108(n). - Petros Hadjicostas, Aug 04 2020

a(n) = ((3 - i*sqrt(3)))/2 - CatalanNumber(n)*hypergeom([1, n + 1/2], [n + 2], 4). - Peter Luschny, Aug 04 2020