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A119259
Central terms of the triangle in A119258.
15
1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889
OFFSET
0,2
COMMENTS
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 06 2022
REFERENCES
R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
LINKS
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, eq (42).
FORMULA
a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
MATHEMATICA
Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
PROG
(Haskell)
a119259 n = a119258 (2 * n) n -- Reinhard Zumkeller, Aug 06 2014
(Python)
from itertools import count, islice
def A119259_gen(): # generator of terms
yield from (1, 3)
a, c = 2, 1
for n in count(1):
yield (a<<n+2)+(1 if n&1 else -1)
a=(3*n+5)*(c:=c*((n<<2)+2)//(n+2))-a>>1
A119259_list = list(islice(A119259_gen(), 20)) # Chai Wah Wu, Apr 26 2023
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, May 11 2006
STATUS
approved