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A064062
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Generalized Catalan numbers C(2; n).
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28
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1, 1, 3, 13, 67, 381, 2307, 14589, 95235, 636925, 4341763, 30056445, 210731011, 1493303293, 10678370307, 76957679613, 558403682307, 4075996839933, 29909606989827, 220510631755773, 1632599134961667, 12133359132082173
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OFFSET
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0,3
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COMMENTS
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a(n+1) = Y_{n}(n+1) = Z_{n}, n >= 0, in the Derrida et al. 1992 reference (see A064094) for alpha=2, beta=1 (or alpha=1, beta=2).
a(n) = number of Dyck n-paths (A000108) in which each upstep (U) not at ground level is colored red (R) or blue (B). For example, a(3)=3 counts URDD, UBDD, UDUD (D=downstep). - David Callan, Mar 30 2007
The sequence a(n)/2^n, with g.f. 1/(1-xc(x)/2), has Hankel transform 1/2^n. - Paul Barry, Apr 14 2008
The REVERT transform of the odd numbers [1,3,5,7,9,...] is [1, -3, 13, -67, 381, -2307, 14589, -95235, 636925, ...] - N. J. A. Sloane, May 26 2017
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LINKS
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FORMULA
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G.f.: (1 + 2*x*C(2*x)) / (1+x) = 1/(1 - x*C(2*x)) with C(x) g.f. of Catalan numbers A000108.
a(n) = A062992(n-1) = Sum_{m = 0..n-1} (n-m)*binomial(n-1+m, m)*(2^m)/n, n >= 1, a(0) = 1.
G.f.: 1/(1-x/(1-2x/(1-2x/(1-2x/(1-.... = 1/(1-x-2x^2/(1-4x-4x^2/(1-4x-4x^2/(1-.... (continued fractions). - Paul Barry, Jan 30 2009
a(n) = (32/Pi)*Integral_{x = 0..1} (8*x)^(n-1)*sqrt(x*(1-x)) / (8*x+1). - Groux Roland, Dec 12 2010
a(n+2) = 8^(n+2)*( c(n+2)-c(1)*c(n+1) - Sum_{i=0..n-1} 8^(-i-2)*c(n-i)*a(i+2) ) with c(n) = Catalan(n+2)/2^(2*n+1). - Groux Roland, Dec 12 2010
a(n) = the upper left term in M^n, M = the production matrix:
1, 1
2, 2, 1
4, 4, 2, 1
8, 8, 4, 2, 1
D-finite with recurrence: n*a(n) + (12-7n)*a(n-1) + 4*(3-2n)*a(n-2) = 0. - R. J. Mathar, Nov 16 2011 (This follows easily from the generating function. - Robert Israel, Nov 30 2014)
G.f. satisfies: A(x) = 1 + A(x)^4 * Integral 1/A(x)^2 dx. - Paul D. Hanna, Dec 24 2013
G.f. satisfies: Integral 1/A(x)^2 dx = x - x^2*G(x), where G(x) is the o.g.f. of A000257, the number of rooted bicubic maps. - Paul D. Hanna, Dec 24 2013
G.f. A(x) satisfies: A(x - 2*x^2) = 1/(1-x). - Paul D. Hanna, Nov 30 2014
a(n) = hypergeometric([1-n, n], [-n], 2) for n > 0. - Peter Luschny, Nov 30 2014
O.g.f. A(x) = 1 + series reversion of (x*(1 - x)/(1 + x)^2). Logarithmically differentiating (A(x) - 1)/x gives 3 + 17*x + 111*x^2 + ..., essentially a g.f for A119259. - Peter Bala, Oct 01 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 6*x^3 + 23*x^4 + ... is a g.f. for A022558.
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. (End)
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MAPLE
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1, seq(simplify(hypergeom([1-n, n], [-n], 2)), n=1..100); # Robert Israel, Nov 30 2014
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MATHEMATICA
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a[0]=1; a[1]=1; a[n_]/; n>=2 := a[n] = a[n-1] + Sum[(a[k] + a[k-1])a[n-k], {k, n-1}]; Table[a[n], {n, 0, 10}] (* David Callan, Aug 27 2009 *)
a[n_] := 2*Sum[ (-1)^j*2^(n-j-1)*Binomial[2*(n-j-1), n-j-1]/(n-j), {j, 0, n-1}] + (-1)^n; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Jul 03 2013 *)
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PROG
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(PARI) {a(n)=polcoeff((3-sqrt(1-8*x+x*O(x^n)))/(2+2*x), n)}
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+A^4*intformal(1/(A^2+x*O(x^n)))); polcoeff(A, n)} \\ Paul D. Hanna, Dec 24 2013
for(n=0, 25, print1(a(n), ", "))
(PARI) {a(n)=polcoeff(1/(1 - serreverse(x-2*x^2 +x^2*O(x^n))), n)}
(Sage)
def a(n):
if n==0: return 1
return hypergeometric([1-n, n], [-n], 2).simplify()
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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STATUS
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approved
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