|
%I #119 Sep 04 2023 11:34:36
%S 1,6,70,924,12870,184756,2704156,40116600,601080390,9075135300,
%T 137846528820,2104098963720,32247603683100,495918532948104,
%U 7648690600760440,118264581564861424,1832624140942590534,28453041475240576740,442512540276836779204,6892620648693261354600
%N a(n) = binomial(4n,2n) or (4*n)!/((2*n)!*(2*n)!).
%C 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
%C 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
%H T. D. Noe, <a href="/A001448/b001448.txt">Table of n, a(n) for n = 0..100</a>
%H R. J. Chapman, T. Y. Chowa, A. Khetana, D. P. Moulton and R. J. Waters, <a href="http://dx.doi.org/10.1016/j.jcta.2008.05.002">Simple formulas for lattice paths avoiding certain periodic staircase boundaries</a>, Journal of Combinatorial Theory, Series A 116 (2009) 205-214.
%H M. Dziemianczuk, <a href="http://arxiv.org/abs/1410.5747">On Directed Lattice Paths With Additional Vertical Steps</a>, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
%H K. H. Pilehrood and T. H. Pilehrood, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Pilehrood/pile5.html">Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients</a>, J. Int. Seq. 18 (2015) 15.11.7.
%H Ricardo A. Podestá, <a href="http://arxiv.org/abs/1603.09156">New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers</a>, arXiv:1603.09156 [math.CO], 2016.
%F 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
%F From _Wolfdieter Lang_, Dec 13 2001: (Start)
%F a(n) = 2*A001700(2*n-1) = (2*n+1)*C(2*n), n >= 1, C(n) := A000108(n) (Catalan).
%F 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)
%F 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
%F a(n) = (1/Pi)*Integral_{x=-2..2} (2+x)^(2*n)/sqrt((2-x)*(2+x)) dx. _Peter Luschny_, Sep 12 2011
%F G.f.: (1/2) * (1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x))). - _Mark van Hoeij_, Oct 25 2011
%F 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
%F 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
%F 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
%F a(n) = hypergeom([1-2*n,-2*n],[2],1)*(2*n+1). - _Peter Luschny_, Sep 22 2014
%F From _Michael Somos_, Oct 22 2014: (Start)
%F 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.
%F 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)
%F 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
%F a(n) = Sum_{i = 0..n} binomial(4*n,i)*binomial(3*n-i-1,n-i). - _Peter Bala_, Sep 29 2015
%F a(n) = A000984(n)*Product_{j=0..n} (2^j/(j!*(2*j-1)!!))*A068424(n, j)^2, with A068424 the falling factorial. See (5.4) in Podestá link. - _Michel Marcus_, Mar 31 2016
%F a(n) = GegenbauerC(2*n, -2*n, -1). - _Peter Luschny_, May 07 2016
%F a(n) = [x^n] 1/sqrt(1 - 4*x)^(2*n+1). - _Ilya Gutkovskiy_, Oct 10 2017
%F 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
%F a(n) = (16^n*(Beta(2*n - 1/2, 1/2) - Beta(2*n - 1/2, 3/2)))/Pi. - _Peter Luschny_, Mar 06 2018
%F E.g.f.: hypergeom([1/4,3/4],[1/2,1],16*x). - _Karol A. Penson_, Mar 08 2018
%F From _Peter Bala_, Feb 16 2020: (Start)
%F a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
%F a(n) = [(x*y)^(2*n)] (1 + x + y)^(4*n). (End)
%F a(n) = (2^n/n!)*Product_{k = n..2*n-1} (2*k + 1). - _Peter Bala_, Feb 26 2023
%e a(n) = (1/Pi)*Integral_{x=0..4} x^(2n)/sqrt(4-(x-2)^2) dx. - _Paul Barry_, Sep 17 2010
%e G.f. = 1 + 6*x + 70*x^2 + 924*x^3 + 12870*x^4 + 184756*x^5 + 2704156*x^6 + ...
%p A001448 := n-> binomial(4*n,2*n) ;
%t Table[Binomial[4n,2n],{n,0,20}] (* _Harvey P. Dale_, Apr 26 2014 *)
%t a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* _Michael Somos_, Oct 22 2014 *)
%o (Magma) [Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // _Vincenzo Librandi_, Sep 13 2011
%o (PARI) a(n)=binomial(4*n,2*n) \\ _Charles R Greathouse IV_, Sep 13 2011
%o (Python)
%o from math import comb
%o def A001448(n): return comb(n<<2,n<<1) # _Chai Wah Wu_, Aug 10 2023
%Y Bisection of A000984. Cf. A002458, A066357, A000984 (k = 0), A091527 (k = 1), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6).
%K nonn,nice,easy
%O 0,2
%A _N. J. A. Sloane_
| 