OFFSET
1,2
COMMENTS
In general, binomial(k*n,n)/k = binomial(k*n-1,n-1).
Sequences in the OEIS related to this identity are:
. C(4n,n) = A005810, C(4n,n)/4 = a(n);
. C(6n,n) = A004355, C(6n,n)/6 is not in the OEIS.
Conjecture: a(n) == 1 (mod n^3) iff n is an odd prime.
It is known that a(p) == 1(mod p^3) for prime p >= 3. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
D. Kruchinin and V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
R. Mestrovic, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
FORMULA
a(n) = binomial(4*n,n)/4 = A005810(n)/4.
a(n) = binomial(4*n-1,n-1).
G.f.: A(x) = B'(x)/B(x), where B(x) = 1 + x*B(x)^4 is g.f. of A002293. - Vladimir Kruchinin, Aug 13 2015
From Peter Bala, Oct 08 2015: (Start)
a(n) = 1/2*[x^n] (C(x)^2)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + ... is the o.g.f. for A002293.
exp( 2*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 9*x^2 + 52*x^3 + ... is the o.g.f. for A069271. (End)
From Peter Bala, Nov 04 2015: (Start)
With an offset of 1, the o.g.f. equals f(x)*g(x)^3, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A257633 (k = 2) and A004331 (k = 4). (End)
a(n) = 1/5*[x^n] (1 + x)/(1 - x)^(3*n + 1) = 1/5*[x^n]( 1/C(-x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A227726. - Peter Bala, Jul 12 2016
a(n) ~ 2^(8*n-3/2)*3^(-3*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
O.g.f.: A(x) = f(x)/(1 - 3*f(x)), where f(x) = series reversion (x/(1 + x)^4) = x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + ... is the o.g.f. of A002293 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/3)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+3)*n+k-1,k) = C(4*n,n)/4 and (1/4)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-4)*n+k-1,k) = C(4*n,n)/4, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(5*n-k-1,2*n-k)*binomial(3*n+k-1,k) = binomial(4*n,n)/4. - Peter Bala, Mar 09 2022
a(n) = [x^n] G(x)^n, where G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the g.f. of A001764. - Peter Bala, Oct 17 2024
EXAMPLE
For n=2, binomial(4*n,n) = binomial(8,2) = 8*7/2 = 28, so a(2) = 28/4 = 7. - Michael B. Porter, Jul 12 2016
MAPLE
seq(binomial(4*n, n)/4, n=1..17);
MATHEMATICA
Table[Binomial[4 n, n]/4, {n, 30}] (* Vincenzo Librandi, Jun 03 2015 *)
PROG
(PARI) a(n) = binomial(4*n, n)/4; /* Joerg Arndt, Apr 02 2013 */
(Magma) [Binomial(4*n, n) div 4: n in [1..25]]; // Vincenzo Librandi, Jun 03 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gary Detlefs, Apr 02 2013
STATUS
approved