A163456 - OEIS

%I #63 May 02 2023 07:49:55

%S 1,9,91,969,10626,118755,1344904,15380937,177232627,2054455634,

%T 23930713170,279871768995,3284214703056,38650751381832,

%U 456002537343216,5391644226101705,63871405575418665,757929628541719755

%N a(n) = binomial(5*n,n)/5.

%C Original name was A163455(n)/4.

%C For prime p, a(p) == 1 (mod p). - _Gary Detlefs_, Aug 03 2013

%C In fact, a(p) == 1 (mod p^3) for prime p >= 5. See Mestrovic, Section 3. - _Peter Bala_, Oct 09 2015

%C From _Robert Israel_, Jul 12 2016: (Start)

%C a(p+1) == 5 (mod p) for primes p >= 5.

%C a(p^(k+1)) == a(p^k) mod p^(3(k+1)) for primes p >= 5. (End)

%D Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994.

%H Robert Israel, <a href="/A163456/b163456.txt">Table of n, a(n) for n = 1..920</a>

%H R. Mestrovic, <a href="http://arxiv.org/abs/1409.3820">Lucas’ theorem: its generalizations, extensions and applications (1878-2014)</a>, arXiv:1409.3820v1 [math.NT], 2014.

%F a(n) = (5*n-1)!/(4*n!*(4*n-1)!).

%F a(n) = binomial(5*n,n)/5. - _Gary Detlefs_, Aug 03 2013

%F From _Peter Bala_, Oct 08 2015: (Start)

%F a(n) = (1/3)*[x^n] (C(x)^3)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274.

%F exp( 3*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 18*x^2 + 136*x^3 + ... is the o.g.f. for A118970. (End)

%F From _Peter Bala_,Jul 12 2016: (Start)

%F a(n) = 1/6*[x^n] (1 + x)/(1 - x)^(4*n + 1).

%F a(n) = 1/6*[x^n] ( 1/C(-x)^6 )^n. Cf. A227726. (End)

%F a(n) ~ 2^(-8*n-3/2)*5^(5*n-1/2)*n^(-1/2)/sqrt(Pi). - _Ilya Gutkovskiy_, Jul 12 2016

%F From _Robert Israel_, Jul 12 2016: (Start)

%F G.f.: x*hypergeom([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125/256)*x).

%F a(n) = 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)/(8*n*(4*n-3)*(2*n-1)*(4*n-1)). (End)

%F O.g.f.: f(x)/(1 - 4*f(x)), where f(x) = series reversion (x/(1 + x)^5) = x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + ... is the o.g.f. of A002294 with the initial term omitted. Cf. A025174. - _Peter Bala_, Feb 03 2022

%F Right-hand side of the identities (1/4)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+4)*n+k-1,k) = C(5*n,n)/5 and (1/5)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-5)*n+k-1,k) = C(5*n,n)/5, both valid for n >= 1 and x arbitrary. - _Peter Bala_, Feb 28 2022

%F Right-hand side of the identity (1/4)*Sum_{k = 0..2*n} (-1)^k*binomial(6*n-k-1,2*n-k)*binomial(4*n+k-1,k) = binomial(5*n,n)/5, for n >= 1. - _Peter Bala_, Mar 09 2022

%F a(n) = (1/2)* [x*n] F(x)^(2*n) = [x^n] G(x)^n for n >= 1, where F(x) = Sum_{k >= 0} 1/(2*k + 1)*binomial(3*k,k)*x^k is the o.g.f. of A001764 and G(x) = Sum_{k >= 0} 1/(3*k + 1)*binomial(4*k,k)*x^k is the o.g.f. of A002293 (apply Concrete Mathematics, equation 5.60, p. 201). - _Peter Bala_, Apr 26 2023

%p seq(binomial(5*n,n)/5, n=1..20); # _Robert Israel_, Jul 12 2016

%t Array[Binomial[5 #, #]/5 &, {18}] (* _Michael De Vlieger_, Oct 09 2015 *)

%o (PARI) a(n) = binomial(5*n,n)/5 \\ _Altug Alkan_, Oct 09 2015

%Y Cf. A000108, A000245, A001764, A002293, A025174, A118970, A163455, A224274, A227726.

%K nonn,easy

%O 1,2

%A _Zak Seidov_, Jul 28 2009

%E Renamed by _Peter Bala_, Oct 08 2015