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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jan 24 2009
STATUS
approved