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A032357 Convolution of Catalan numbers and powers of -1. 8
1, 0, 2, 3, 11, 31, 101, 328, 1102, 3760, 13036, 45750, 162262, 580638, 2093802, 7601043, 27756627, 101888163, 375750537, 1391512653, 5172607767, 19293659253, 72188904387, 270870709263, 1019033438061, 3842912963391, 14524440108761 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Absolute value of the alternating sum of Catalan Numbers. - Alexander Adamchuk, Jul 03 2006

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

Hankel transform is F(2n+1). - Paul Barry, Jul 22 2008

Equals INVERTi transform of A000958. [Gary W. Adamson, Apr 10 2009]

Inverse binomial transform of A002212 . [Philippe Deléham, Sep 17 2009]

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

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

S. B. Ekhad, M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017)

FORMULA

G.f.: c(x)/(1+x), where c(x) is the g.f. for Catalan numbers.

a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k), where C(k)=A000108(k).

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

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

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

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

a(n) = (-1)^n/A001622 + A000108(n+1)*hypergeom([1, n+3/2], [n+3], -4). - Vladimir Reshetnikov, Oct 02 2016

a(n) ~ 2^(2*n+2) / (5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 03 2016

a(n) = (A000108(n) * (2 + (n+1)*hypergeom([1,-n], [1/2], 5/4)) - (-1)^n)/2. - Vladimir Reshetnikov, Oct 03 2016

MAPLE

rec:= (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2)=0:

A:= gfun:-rectoproc({rec, a(0)=1, a(1)=0}, a(n), remember):

seq(A(n), n=0..50); # Robert Israel, May 22 2015

MATHEMATICA

Table[Sum[(-1)^(k+n)*CatalanNumber[k], {k, 0, n}], {n, 0, 60}] (* Alexander Adamchuk, Jul 03 2006 *)

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 *)

Table[(CatalanNumber[n] (2 + (n + 1) Hypergeometric2F1[1, -n, 1/2, 5/4]) - (-1)^n)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

PROG

(Sage)

def A032357():

    f, c, n = 1, 1, 1

    while True:

        yield f

        n += 1

        c = c * (4*n - 6) // n

        f = c - f

a = A032357()

print [a.next() for _ in range(27)] # Peter Luschny, Nov 30 2016

CROSSREFS

Cf. A000108, A000958, A014137, A014138, A033297, A045468, A038872, A064739.

Sequence in context: A191058 A080155 A235625 * A144056 A268687 A062630

Adjacent sequences:  A032354 A032355 A032356 * A032358 A032359 A032360

KEYWORD

easy,nonn

AUTHOR

Wolfdieter Lang

EXTENSIONS

More terms from Christian G. Bower, Apr 15 1998

More terms from Alexander Adamchuk, Jul 03 2006

STATUS

approved

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Last modified December 12 01:08 EST 2017. Contains 295936 sequences.