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A001448
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a(n) = binomial(4n,2n) or (4*n)!/((2*n)!*(2*n)!).
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37
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1, 6, 70, 924, 12870, 184756, 2704156, 40116600, 601080390, 9075135300, 137846528820, 2104098963720, 32247603683100, 495918532948104, 7648690600760440, 118264581564861424, 1832624140942590534, 28453041475240576740, 442512540276836779204, 6892620648693261354600
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OFFSET
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0,2
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COMMENTS
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Corollary 8 in Chapman et alia says: "For n>=1, there are binomial(4n,2n) binary sequences of length 4n+1 with the property that for all j, the j-th occurrence of 10 appears in positions 4j+1 and 4j+2 or later (if it exists at all)." - Peter Luschny, Nov 21 2011
Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 2. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015
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LINKS
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FORMULA
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Using Stirling's formula in sequence A000142 it is easy to get the asymptotic expression a(n) ~ 16^n / sqrt(2 * Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
a(n) = 2*A001700(2*n-1) = (2*n+1)*C(2*n), n >= 1, C(n) := A000108(n) (Catalan).
G.f.: (1-y*((1+4*y)*c(y)-(1-4*y)*c(-y)))/(1-(4*y)^2) with y^2=x, c(y) = g.f. for A000108 (Catalan). (End)
a(n) ~ 2^(-1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - (1/16)*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = (1/Pi)*Integral_{x=-2..2} (2+x)^(2*n)/sqrt((2-x)*(2+x)) dx. Peter Luschny, Sep 12 2011
G.f.: (1/2) * (1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x))). - Mark van Hoeij, Oct 25 2011
Sum_{n>=1} 1/a(n) = 16/15 + Pi*sqrt(3)/27 - 2*sqrt(5)*log(phi)/25, [T. Trif, Fib Quart 38 (2000) 79] with phi=A001622. - R. J. Mathar, Jul 18 2012
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Dec 02 2012
G.f.: sqrt((1 + sqrt(1-16*x))/(2*(1-16*x))) = 1 + 6*x/(G(0)-6*x), where G(k) = 2*x*(4*k+3)*(4*k+1) + (2*k+1)*(k+1) - 2*x*(k+1)*(2*k+1)*(4*k+5)*(4*k+7)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 30 2013
a(n) = hypergeom([1-2*n,-2*n],[2],1)*(2*n+1). - Peter Luschny, Sep 22 2014
0 = a(n)*(+65536*a(n+2) - 16896*a(n+3) + 858*a(n+4)) + a(n+1)*(-3584*a(n+2) + 1176*a(n+3) - 66*a(n+4)) + a(n+2)*(+14*a(n+2) - 14*a(n+3) + a(n+4)) for all n in Z.
0 = a(n)^2*(+196608*a(n+1)^2 - 40960*a(n+1)*a(n+2) + 2100*a(n+2)^2) + a(n)*a(n+1)*(-12288*a(n+1)^2 + 2840*a(n+1)*a(n+2) - 160*a(n+2)^2) + a(n+1)^2*(+180*a(n+1)^2 - 48*a(n+1)*a(n+2) + 3*a(n+2)^2) for all n in Z. (End)
a(n) = [x^n] ( (1 + x)^4/(1 - x)^2 )^n; exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... = Sum_{n >= 0} A066357(n+1)*x^n. - Peter Bala, Jun 23 2015
a(n) = Sum_{i = 0..n} binomial(4*n,i)*binomial(3*n-i-1,n-i). - Peter Bala, Sep 29 2015
a(n) is the n-th moment of the positive weight function w(x) on (0,16), i.e. in Maple notation, a(n) = int(x^n*w(x), x = 0..16), n = 0,1,..., where w(x) = (1/(2*Pi))/((sqrt(4 - sqrt(x))*x^(3/4)). The function w(x) is the solution of the Hausdorff moment problem and is unique. - Karol A. Penson, Mar 06 2018
a(n) = (16^n*(Beta(2*n - 1/2, 1/2) - Beta(2*n - 1/2, 3/2)))/Pi. - Peter Luschny, Mar 06 2018
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(2*n)] (1 + x + y)^(4*n). (End)
a(n) = (2^n/n!)*Product_{k = n..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
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EXAMPLE
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a(n) = (1/Pi)*Integral_{x=0..4} x^(2n)/sqrt(4-(x-2)^2) dx. - Paul Barry, Sep 17 2010
G.f. = 1 + 6*x + 70*x^2 + 924*x^3 + 12870*x^4 + 184756*x^5 + 2704156*x^6 + ...
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MAPLE
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A001448 := n-> binomial(4*n, 2*n) ;
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MATHEMATICA
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a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* Michael Somos, Oct 22 2014 *)
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PROG
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(Magma) [Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011
(Python)
from math import comb
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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