%I M3625 #151 Aug 01 2023 09:55:38

%S 1,4,28,220,1820,15504,134596,1184040,10518300,94143280,847660528,

%T 7669339132,69668534468,635013559600,5804731963800,53194089192720,

%U 488526937079580,4495151581425648,41432089765583440,382460951663844400

%N a(n) = binomial(4n,n).

%C Start off with 0 balls in a box. Find the number of ways you can throw 3 balls back out. Then continue to throw 4 balls into the box after each stage. (I.e., the first stage is 0. Then at the next stage there are 4 ways to throw 3 balls back out.) - Ruppi Rana (ruppirana007(AT)hotmail.com), Mar 03 2004

%C Central coefficients of A094527. - _Paul Barry_, Mar 08 2011

%C This is the case m = 2n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - _Bruno Berselli_, Apr 27 2012

%C A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. - _Tom Copeland_, Oct 10 2012

%C Conjecture: a(n) == 4 (mod n^3) iff n is prime. - _Gary Detlefs_, Apr 03 2013

%C For prime p, the congruence a(p) = binomial(4*p,p) = 4 (mod p^3) is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - _Peter Bala_, Dec 28 2014

%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

%D Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Vaclav Kotesovec, <a href="/A005810/b005810.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..100 from T. D. Noe, terms 101..213 from Muniru A Asiru)

%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

%H Tom Copeland, <a href="https://tcjpn.files.wordpress.com/2013/04/discrdeltas9-6-20122.pdf">Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers</a>, 2012, pp. 5-6.

%H R. Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

%H Ruppi Rana, <a href="http://web.njit.edu/~rr93/p12.htm">Title?</a> [Broken link]

%F a(n) is asymptotic to c*(256/27)^n/sqrt(n) with c = sqrt(2 / (3 Pi)) = 0.460658865961780639... - _Benoit Cloitre_, Jan 26 2003; corrected by _Charles R Greathouse IV_, Dec 14 2006

%F a(n) = Sum_{k=0..2n} binomial(2n,k)*binomial(2n,k-n). - _Paul Barry_, Mar 08 2011

%F G.f.: g/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - _Mark van Hoeij_, Nov 11 2011

%F D-finite with recurrence: 3*n*(3*n-1)*(3*n-2)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - _R. J. Mathar_, Dec 02 2012

%F a(n) = binomial(4*n,n-1)*(3*n+1)/n. - _Gary Detlefs_, Apr 03 2013

%F a(n) = C(4*n-1,n-1)*C(16*n^2,2)/(3*n*C(4*n+1,3)), n>0. - _Gary Detlefs_, Jan 02 2014

%F a(n) = Sum_{i,j,k = 0..n} binomial(n,i)*binomial(n,j)*binomial(n,k)* binomial(n,i+j+k). - _Peter Bala_, Dec 28 2014

%F a(n) = GegenbauerC(n, -2*n, -1). - _Peter Luschny_, May 07 2016

%F From _Ilya Gutkovskiy_, Nov 22 2016: (Start)

%F O.g.f.: 3F2(1/4,1/2,3/4; 1/3,2/3; 256*x/27).

%F E.g.f.: 3F3(1/4,1/2,3/4; 1/3,2/3,1; 256*x/27). (End)

%F a(n) = hypergeom([-3*n, -1*n], [1], 1). - _Peter Luschny_, Mar 19 2018

%F RHS of the identity Sum_{k = 0..2*n} (-1)^(n+k)*binomial(4*n, k)* binomial(4*n, 2*n-k) = binomial(4*n,n). - _Peter Bala_, Oct 07 2021

%F From _Peter Bala_, Feb 20 2022: (Start)

%F The o.g.f. A(x) satisfies the differential equation

%F (-256*x^3 + 27*x^2)*A(x)''' + (-1152*x^2 + 54*x)*A(x)'' + (-816*x + 6)*A(x)' - 24*A(x) = 0 with A(0) = 1, A'(0) = 4 and A''(0) = 56.

%F Algebraic equation: (1 - A(x))*(1 + 3*A(x))^3 + 256*x*A(x)^4 = 0.

%F Sum_{n >= 1} a(n)*( x*(3*x + 4)^3/(256*(1 + x)^4) )^n = x. (End)

%e G.f. = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + 15504*x^5 + 134596*x^6 + ...

%p seq(binomial(4*n,n),n=0..20); # _Muniru A Asiru_, Mar 19 2018

%t Table[Binomial[4n,n],{n,0,19}] (* _Geoffrey Critzer_, Sep 15 2013 *)

%o (Magma) [ Binomial(4*n,n): n in [0..100] ]; // _Vincenzo Librandi_, Apr 13 2011

%o (Haskell)

%o a005810 n = a007318 (4*n) n -- _Reinhard Zumkeller_, Mar 04 2012

%o (PARI) a(n) = binomial(4*n, n); \\ _Altug Alkan_, Mar 19 2018

%o (GAP) List([0..20],n->Binomial(4*n,n)); # _Muniru A Asiru_, Mar 19 2018

%o (Python)

%o from math import comb

%o def A005810(n): return comb(n<<2,n) # _Chai Wah Wu_, Aug 01 2023

%Y binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

%Y Cf. A007318, A182400, A262261.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Henry Bottomley_, Oct 06 2000

%E Corrected by _T. D. Noe_, Jan 16 2007