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

0,2

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.

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

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

Prog

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

Crossrefs

Cf. A000108, A014138, A057079, A155586.

Partial sums of A120588.

Sequence in context: A099967 A106531 A101163 * A125658 A001200 A050837

Adjacent sequences: A155584 A155585 A155586 * A155588 A155589 A155590

Keyword

easy,nonn

Author

Paul Barry, Jan 24 2009

Status

approved