|
%I #39 Jun 24 2022 23:37:36
%S 1,2,3,5,10,24,66,198,627,2057,6919,23715,82501,290513,1033413,
%T 3707853,13402698,48760368,178405158,656043858,2423307048,8987427468,
%U 33453694488,124936258128,467995871778,1757900019102,6619846420554
%N Expansion of (1 + x*c(x))/(1 - x), where c(x) is the g.f. of A000108.
%C Row sums of A155586.
%C Hankel transform is A057079(n+2).
%C From _Petros Hadjicostas_, Aug 03 2020: (Start)
%C 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).
%C 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)
%H Reinhard Zumkeller, <a href="/A155587/b155587.txt">Table of n, a(n) for n = 0..1000</a>
%H Gerard Cohen and Jean-Pierre Flori, <a href="https://eprint.iacr.org/2011/400">On a generalized combinatorial conjecture involving addition mod 2^k - 1</a>, IACR, Report 2011/400.
%H Jean-Pierre Flori, <a href="https://perso.telecom-paristech.fr/flori/thesis/thesis.pdf">Fonctions booléennes, courbes algébriques et multiplication complexe</a>, Thesis, ParisTech, Feb 03 2012.
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%F a(n) = 1 + Sum_{k=0..n-1} A000108(k).
%F Conjecture: n*a(n) + (6-5*n)*a(n-1) + 2*(2*n-3)*a(n-2) = 0. - _R. J. Mathar_, Nov 15 2011
%F a(n) = A014138(n-1) + 2 for n > 0. - _Reinhard Zumkeller_, Mar 01 2013 [Corrected by _Petros Hadjicostas_, Aug 03 2020]
%F a(n+1) - a(n) = A000108(n). - _Petros Hadjicostas_, Aug 04 2020
%F a(n) = ((3 - i*sqrt(3)))/2 - CatalanNumber(n)*hypergeom([1, n + 1/2], [n + 2], 4). - _Peter Luschny_, Aug 04 2020
%p CatalanNumber := n -> binomial(2*n, n)/(n+1):
%p a := n -> ((3 - I*sqrt(3)))/2 - CatalanNumber(n)*hypergeom([1, n+1/2], [n+2], 4):
%p seq(simplify(a(n)), n=0..26); # _Peter Luschny_, Aug 04 2020
%o (Haskell)
%o a155587 n = a155587_list !! n
%o a155587_list = scanl (+) 1 a000108_list -- _Reinhard Zumkeller_, Mar 01 2013
%Y Cf. A000108, A014138, A057079, A155586.
%Y Partial sums of A120588.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jan 24 2009
| 