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 A001448 a(n) = binomial(4n,2n) or (4*n)!/((2*n)!*(2*n)!). 37
 1, 6, 70, 924, 12870, 184756, 2704156, 40116600, 601080390, 9075135300, 137846528820, 2104098963720, 32247603683100, 495918532948104, 7648690600760440, 118264581564861424, 1832624140942590534, 28453041475240576740, 442512540276836779204, 6892620648693261354600 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 LINKS T. D. Noe, Table of n, a(n) for n = 0..100 R. J. Chapman, T. Y. Chowa, A. Khetana, D. P. Moulton and R. J. Waters, Simple formulas for lattice paths avoiding certain periodic staircase boundaries, Journal of Combinatorial Theory, Series A 116 (2009) 205-214. M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014. K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7. Ricardo A. Podestá, New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers, arXiv:1603.09156 [math.CO], 2016. FORMULA 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 From Wolfdieter Lang, Dec 13 2001: (Start) 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 From Michael Somos, Oct 22 2014: (Start) 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) = 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 a(n) = GegenbauerC(2*n, -2*n, -1). - Peter Luschny, May 07 2016 a(n) = [x^n] 1/sqrt(1 - 4*x)^(2*n+1). - Ilya Gutkovskiy, Oct 10 2017 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 E.g.f.: hypergeom([1/4,3/4],[1/2,1],16*x). - Karol A. Penson, Mar 08 2018 From Peter Bala, Feb 16 2020: (Start) 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 EXAMPLE 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 + ... MAPLE A001448 := n-> binomial(4*n, 2*n) ; MATHEMATICA Table[Binomial[4n, 2n], {n, 0, 20}] (* Harvey P. Dale, Apr 26 2014 *) a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* Michael Somos, Oct 22 2014 *) PROG (Magma) [Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011 (PARI) a(n)=binomial(4*n, 2*n) \\ Charles R Greathouse IV, Sep 13 2011 (Python) from math import comb def A001448(n): return comb(n<<2, n<<1) # Chai Wah Wu, Aug 10 2023 CROSSREFS Bisection of A000984. Cf. A002458, A066357, A000984 (k = 0), A091527 (k = 1), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6). Sequence in context: A286527 A104900 A186667 * A024489 A354328 A036361 Adjacent sequences: A001445 A001446 A001447 * A001449 A001450 A001451 KEYWORD nonn,nice,easy AUTHOR N. J. A. Sloane STATUS approved

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Last modified December 2 07:23 EST 2023. Contains 367510 sequences. (Running on oeis4.)