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 A032357 Convolution of Catalan numbers and powers of -1. 8

%I

%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(n) Primes congruent to {1, 4} mod 5. Prime p divides a(2p+1) for p = 5, 11, 19, 29, 31, 41, 59, 61, 71,... = A038872(n) 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(2n+1). - _Paul Barry_, Jul 22 2008

%C Equals INVERTi transform of A000958. [_Gary W. Adamson_, Apr 10 2009]

%C Inverse binomial transform of A002212 . [_Philippe DelĂ©ham_, 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 S. B. Ekhad, M. 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 Catalan numbers.

%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((-5)^k*binomial(n+1, k)/binomial(2*k, k), k=0..n+1))/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 [a.next() for _ in range(27)] # _Peter Luschny_, Nov 30 2016

%Y Cf. A000108, A000958, A014137, A014138, A033297, A045468, A038872, 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

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Last modified October 20 23:39 EDT 2018. Contains 316405 sequences. (Running on oeis4.)