|
| |
|
|
A224274
|
|
a(n) = binomial(4*n,n)/4.
|
|
17
|
|
|
|
1, 7, 55, 455, 3876, 33649, 296010, 2629575, 23535820, 211915132, 1917334783, 17417133617, 158753389900, 1451182990950, 13298522298180, 122131734269895, 1123787895356412, 10358022441395860, 95615237915961100, 883829035553043580, 8179808679272664720, 75788358475481302185
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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(2n,n) = A000984, C(2n,n)/2 = A001700;
. C(3n,n) = A005809, C(3n,n)/3 = A025174;
. C(4n,n) = A005810, C(4n,n)/4 = a(n);
. C(5n,n) = A001449, C(5n,n)/5 = A163456;
. 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.
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
|
|
|
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
|
Cf. A001700, A025174, A163456, A002293, A069271, A004331, A005810, A052203, A257633, A262977, A227726.
Sequence in context: A083068 A097189 A049028 * A096951 A113714 A246459
Adjacent sequences: A224271 A224272 A224273 * A224275 A224276 A224277
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Gary Detlefs, Apr 02 2013
|
|
|
STATUS
|
approved
|
| |
|
|
