|
|
A005810
|
|
a(n) = binomial(4n,n).
(Formerly M3625)
|
|
47
|
|
|
1, 4, 28, 220, 1820, 15504, 134596, 1184040, 10518300, 94143280, 847660528, 7669339132, 69668534468, 635013559600, 5804731963800, 53194089192720, 488526937079580, 4495151581425648, 41432089765583440, 382460951663844400
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
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
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
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
Conjecture: a(n) == 4 (mod n^3) iff n is prime. - Gary Detlefs, Apr 03 2013
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
|
|
REFERENCES
|
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.
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.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
LINKS
|
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Ruppi Rana, Title? [Broken link]
|
|
FORMULA
|
a(n) = Sum_{k=0..2n} binomial(2n,k)*binomial(2n,k-n). - Paul Barry, Mar 08 2011
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
a(n) = binomial(4*n,n-1)*(3*n+1)/n. - Gary Detlefs, Apr 03 2013
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
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
O.g.f.: 3F2(1/4,1/2,3/4; 1/3,2/3; 256*x/27).
E.g.f.: 3F3(1/4,1/2,3/4; 1/3,2/3,1; 256*x/27). (End)
a(n) = hypergeom([-3*n, -1*n], [1], 1). - Peter Luschny, Mar 19 2018
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
The o.g.f. A(x) satisfies the differential equation
(-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.
Algebraic equation: (1 - A(x))*(1 + 3*A(x))^3 + 256*x*A(x)^4 = 0.
Sum_{n >= 1} a(n)*( x*(3*x + 4)^3/(256*(1 + x)^4) )^n = x. (End)
|
|
EXAMPLE
|
G.f. = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + 15504*x^5 + 134596*x^6 + ...
|
|
MAPLE
|
|
|
MATHEMATICA
|
|
|
PROG
|
(Haskell)
(PARI) a(n) = binomial(4*n, n); \\ Altug Alkan, Mar 19 2018
(GAP) List([0..20], n->Binomial(4*n, n)); # Muniru A Asiru, Mar 19 2018
(Python)
from math import comb
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|