%I #80 Feb 05 2022 15:53:57
%S 1,0,2,3,11,31,101,328,1102,3760,13036,45750,162262,580638,2093802,
%T 7601043,27756627,101888163,375750537,1391512653,5172607767,
%U 19293659253,72188904387,270870709263,1019033438061,3842912963391,14524440108761
%N Convolution of Catalan numbers and powers of -1.
%C Absolute value of the alternating sum of Catalan Numbers. - _Alexander Adamchuk_, Jul 03 2006
%C Sums of two consecutive terms are a(n-1) + a(n) = 1, 2, 5, 14, 42, ... = A000108(n) (Catalan Numbers). The prime p divides a((p-3)/2) for p = 11, 19, 29, 31, 41, 59, 61, 71, ... = A045468 (Primes congruent to {1, 4} mod 5). Prime p divides a(2*p+1) for p = 5, 11, 19, 29, 31, 41, 59, 61, 71, ... = A038872 (Primes congruent to {0, 1, 4} mod 5). Also odd primes where 5 is a square mod p. - _Alexander Adamchuk_, Jul 03 2006
%C Hankel transform is F(2*n+1), where F = A000045. - _Paul Barry_, Jul 22 2008
%C Equals INVERTi transform of A000958. - _Gary W. Adamson_, Apr 10 2009
%C Inverse binomial transform of A002212. - _Philippe Deléham_, Sep 17 2009
%C Number of singleton and plus-decomposable (2143, 2413, 3142)-avoiding permutations with no +bonds (ascents by 1), with offset 1. Equivalently, number of (2143, 2413, 3142)-avoiding permutations that start with 1 or end with n (top entry). E.g., 132 and 213 for n = 3; 1324, 1432, 3214 for n = 4. - _Alexander Burstein_, May 22 2015
%H Vincenzo Librandi, <a href="/A032357/b032357.txt">Table of n, a(n) for n = 0..1000</a>
%H Shalosh B. Ekhad and Mingjia Yang, <a href="http://sites.math.rutgers.edu/~zeilberg/tokhniot/oMathar1maple12.txt">Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences</a>, 2017.
%F G.f.: c(x)/(1 + x), where c(x) is the g.f. for the Catalan numbers A000108.
%F a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k), where C(k) = A000108(k).
%F a(n) = ((-1)^(n+1) - binomial(2*(n+1), n+1)*Sum_{k=0..n+1} (-5)^k*binomial(n+1, k)/binomial(2*k, k))/2.
%F a(n) = C(2*n, n)/(n+1) - a(n-1) = A000108(n) - a(n-1) with a(0) = 1. - _Labos Elemer_, Apr 26 2003
%F Conjecture: (n+1)*a(n) + 3*(-n+1)*a(n-1) + 2*(-2*n+1)*a(n-2) = 0. - _R. J. Mathar_, Nov 30 2012
%F Conjecture is true since the g.f. satisfies (x - 3*x^2 - 4*x^3)*g'(x) + (1 - 6*x^2)*g(x) = 1. - _Robert Israel_, May 22 2015
%F a(n) = (-1)^n/A001622 + A000108(n+1)*hypergeom([1, n + 3/2], [n + 3], -4). - _Vladimir Reshetnikov_, Oct 02 2016
%F a(n) ~ 2^(2*n + 2) / (5*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 03 2016
%F a(n) = (A000108(n) * (2 + (n + 1)*hypergeom([1,-n], [1/2], 5/4)) - (-1)^n)/2. - _Vladimir Reshetnikov_, Oct 03 2016
%p rec:= (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2)=0:
%p A:= gfun:-rectoproc({rec,a(0)=1,a(1)=0},a(n),remember):
%p seq(A(n),n=0..50); # _Robert Israel_, May 22 2015
%t Table[Sum[(-1)^(k+n)*CatalanNumber[k],{k,0,n}],{n,0,60}] (* _Alexander Adamchuk_, Jul 03 2006 *)
%t Round@Table[(-1)^n/GoldenRatio + CatalanNumber[n + 1] Hypergeometric2F1[1, n + 3/2, n + 3, -4], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - _Vladimir Reshetnikov_, Oct 02 2016 *)
%t Table[(CatalanNumber[n] (2 + (n + 1) Hypergeometric2F1[1, -n, 1/2, 5/4]) - (-1)^n)/2, {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 03 2016 *)
%o (Sage)
%o def A032357():
%o f, c, n = 1, 1, 1
%o while True:
%o yield f
%o n += 1
%o c = c * (4*n - 6) // n
%o f = c - f
%o a = A032357()
%o print([next(a) for _ in range(27)]) # _Peter Luschny_, Nov 30 2016
%Y Cf. A000045, A000108, A000958, A001622, A002212, A014137, A014138, A033297, A038872, A045468, A064739.
%K easy,nonn
%O 0,3
%A _Wolfdieter Lang_
%E More terms from _Christian G. Bower_, Apr 15 1998
%E More terms from _Alexander Adamchuk_, Jul 03 2006