cp's OEIS Frontend

a(n) = Sum_{i=0..n} 1/(5*n+i+1) * C(5*n+1, n-i) * C(5*n+2*i, i).

a(n) = Sum_{i=0..2*n} (-1)^i/(5*i+1) * C((5*i+1)/2, i) * 1/(1+5*(2*n-i)) * C((1+5*(2*n-i))/2, 2*n-i).

G.f. A(z) satisfies: A(z) = 1+2*z*A^5-z*A^6+z*A^7+z^2*A^10. [Corrected by Bryan T. Ek, Oct 30 2017]

G.f.: A(z) = exp(C(5,2)*z/5 + C(10,4)*z^2/10 + C(15,6)*z^3/15 + ...). - Don Knuth, Oct 05 2014

Recurrence: 216*(n-1)*n*(2*n-1)*(3*n-4)*(3*n-2)*(3*n-1)*(3*n+1)*(6*n-1)*(6*n+1)*(5625*n^4 - 38550*n^3 + 97425*n^2 - 107784*n + 44044)*a(n) = 540*(n-1)*(3*n-4)*(3*n-2)*(126562500*n^10 - 1373625000*n^9 + 6557484375*n^8 - 18192221250*n^7 + 32549973750*n^6 - 39248008800*n^5 + 32203028675*n^4 - 17641491134*n^3 + 6113558828*n^2 - 1191132600*n + 96112128)*a(n-1) - 450*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(63281250*n^9 - 718453125*n^8 + 3556125000*n^7 - 10046426250*n^6 + 17765816250*n^5 - 20240090325*n^4 + 14698993900*n^3 - 6468702396*n^2 + 1533535184*n - 142988160)*a(n-2) + 78125*(n-2)*(5*n-14)*(5*n-13)*(5*n-12)*(5*n-11)*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(5625*n^4 - 16050*n^3 + 15525*n^2 - 6084*n + 760)*a(n-3). - Vaclav Kotesovec, Oct 05 2014

Asymptotics (Duchon, 2000): a(n) ~ c * (3125/108)^n / n^(3/2), where c = 0.0876612192439026461763141944768209255550234422281635788... (constant corrected, in the reference "On the enumeration and generation of generalized Dyck words", p.132 is a wrong value 0.0887). - Vaclav Kotesovec, Oct 05 2014, c = sqrt(5*(10^(2/3) - 5^(1/3)/2^(2/3) - 2))/(18*sqrt(Pi)). - Vaclav Kotesovec, Sep 16 2021

a(n) = Gamma(n+4/5)*Gamma(n+3/5)*Gamma(n+2/5)*3125^n*hypergeom([-n, (5/2)*n+1, (5/2)*n+1/2], [5*n+2, 4*n+2], -4)*Gamma(n+1/5)/ (Pi^2*csc((2/5)*Pi)*csc((1/5)*Pi)*Gamma(4*n+2)). - Robert Israel, Oct 05 2014

a(n) = A002294(n)*hypergeom([-n,5*n/2+1/2,5*n/2+1],[4*n+2,5*n+2],-4). - Peter Luschny, Oct 05 2014

O.g.f. A(x) satisfies: A(x)^5 = 1/x*series reversion( x/((1+x)*C(x))^5 ), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. See A001450. - Peter Bala, Oct 05 2015

The sequence defined by b(n) := [x^n] A(x)^n begins [1, 2, 50, 1415, 42258, 1300727, 40820837, 1298493730, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 7 (checked up to p = 101). [Added 23 Oct 2024: More generally, let r be an integer and s a positive integer and define a sequence u(n) by u(n) = [x^(s*n)] A(x)^(r*n). Then we conjecture that the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 7 and positive integers n and k.] - Peter Bala, Sep 12 2021

Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for primes p >= 7 and positive integers n and k. - Peter Bala, Oct 24 2024

a(n) = A001222(A001450(n)). - Michel Marcus, Nov 09 2017

a(n) = A000984(3*n).

a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} binomial(n, i)*binomial(n, j) *binomial(n, k)*binomial(3n, i+j+k). - Benoit Cloitre, Mar 08 2005

O.g.f. (with a(0):=1): (cb(x^(1/3)) + sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))+1+2*x^(1/3)))/3, with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x) = 1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials). - Wolfdieter Lang, Mar 24 2011

D-finite with recurrence n*(3n-1)*(3n-2)*a(n) = 8*(6n-5)*(6n-1)*(2n-1)*a(n-1). - R. J. Mathar, Sep 17 2012

a(n) = GegenbauerC(3*n, -3*n, -1). - Peter Luschny, May 07 2016

a(n) = hypergeom([-3*n, -3*n], [1], 1). - Peter Luschny, Mar 19 2018

a(n) ~ 2^(6*n)/sqrt(3*Pi*n). - Vaclav Kotesovec, Jun 07 2019

From Peter Bala, Feb 16 2020: (Start)

a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.

a(n) = [(x*y)^(3*n)] (1 + x + y)^(6*n). Cf. A001448. (End)

Conjecture: a(n) = [x^n] G(x)^(2*n), where G(x) = (1 + x)*(1 - 6*x + x^2)/(2*x) + (x^2 - 1)*sqrt(1 - 14*x + x^2)/(2*x) = 1 + 10*x + 81*x^2 + 720*x^3 + .... The algebraic function G(x) satisfies the quadratic equation x*G(x)^2 - (1 - 5*x - 5*x^2 + x^3)*G(x) + (1 + x)^4 = 0. Cf. A001450. - Peter Bala, Oct 27 2022

a(n) = Sum_{k = 0..3*n} binomial(3*n+k-1, k). - Peter Bala, Jun 04 2024

O.g.f: 3F2(1/6,1/2,5/6; 1/3,2/3 ; 64*x). - R. J. Mathar, Jan 11 2025

a(n) = Sum_{k=0..n} A000108(k)*C(1,n-k).

a(0)= 1, a(n) = A005807(n-1) for n>0. - Philippe Deléham, Nov 25 2009

(n+1)*a(n) +(-3*n+1)*a(n-1) +2*(-2*n+5)*a(n-2)=0, for n>2. - R. J. Mathar, Feb 10 2015

-(n+1)*(5*n-6)*a(n) +2*(5*n-1)*(2*n-3)*a(n-1)=0. - R. J. Mathar, Feb 10 2015

The o.g.f. A(x) satisfies [x^n] A(x)^(5*n) = binomial(5*n,2*n) = A001450(n). Cf. A182959. - Peter Bala, Oct 04 2015

Let F(x) be the g.f. of A182960, then g.f. of this sequence satisfies:

* A(x) = F(x/A(x)^3) and A(x*F(x)^3) = F(x);

* A(x) = [x/Series_Reversion( x*F(x)^3 )]^(1/3).

G.f.: 1/2/x - 1/2 - x - (1+x)/x/G(0), where G(k)= 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013

a(n) ~ 9*2^(3*n-2)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 29 2013

From Peter Bala, Oct 04 2015: (Start)

O.g.f. A(x) = (1 + x)*(2*C(2*x) - 1), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.

[x^n] A(x)^(3*n) = binomial(6*n,2*n). Cf. with the identity [x^n] ( (1 + x)*C(x) )^(5*n) = binomial(5*n,2*n) = A001450(n). (End)

Conjecture: D-finite with recurrence (n+1)*a(n) +(-7*n+3)*a(n-1) +4*(-2*n+5)*a(n-2)=0. - R. J. Mathar, Jan 22 2020

From Peter Bala, May 15 2023: (Start)

a(n) = 3*(2^n)*(3*n - 1)/(n*(n + 1)) * binomial(2*n-2,n-1) for n >= 2.

(n + 1)*(3*n - 4)*a(n) = 4*(2*n - 3)*(3*n - 1)*a(n-1) for n >= 3 with a(2) = 20. Mathar's conjectured second order recurrence above follows from this. (End)

[x^n] A(x)^n = A372215(n). - Peter Bala, Nov 07 2024

a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(4*n, 2*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity, Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).

P-recursive: a(n) = (20/9)*(4*n-1)*(4*n-3)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)/((3*n-1)^2*(3*n-2)^2*n^2) * a(n-1) with a(0) = 1.

a(n) ~ c^n * sqrt(10)/(6*Pi*n), where c = (2^6)*(5^5)/(3^6).

a(n) = [x^n] G(x)^(20*n), where the power series G(x) = 1 + 2*x + 69*x^2 + 5647*x^3 + 618860*x^4 + 79241349*x^5 + 11177111981*x^6 + 1684171189810*x^7 + 266238907746252*x^8 + ... appears to have integer coefficients.

exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^20, where the power series F(x) = 1 + 2*x + 149*x^2 + 18647*x^3 + 2913620*x^4 + 515276389*x^5 + 98628630997*x^6 + 19944410220744*x^7 + 4199273746072180*x^8 + ... appears to have integer coefficients.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r [added Aug 04 2023: the conjecture follows from Meštrović. equation 39].

a(n) = binomial(4*n,n)*binomial(5*n,2*n). - Christian Krause, Aug 03 2023

G.f.: A(x) = 1 + (x*B(x)')/(B(x)) where B(x) = 2 * (1 + x*B(x)^2)^2 / (1 - 2*x*B(x)^2 + sqrt(1-8*x*B(x)^2)).

a(n) ~ 3^(6*n-1/2) / (sqrt(Pi*n) * 2^(4*n+3/2)). - Vaclav Kotesovec, Jul 01 2015

a(n) = A025174(2*n), n>0. - R. J. Mathar, Jun 07 2016

From Peter Bala, Jun 08 2024: (Start)

a(n) = (9/2)*(6*n-1)*(6*n-5)*(3*n-1)*(3*n-2)/((4*n-1)*(4*n-3)*(2*n-1)*n) * a(n-1) with a(0) = 1 and a(1) = 5.

Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(5*n-k, 2*n-k) = (1/3)*binomial(6*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(5*n-k, 2*n-k) = binomial(4*n, 2*n). (End)

From Karol A. Penson, Jan 26 2025: (Start)

G.f. for 3*a(n),a(0)=1, denoted A, expressible entirely by radicals: A = A1 + A2 with

A1 = ((4*sqrt(4 - 27*sqrt(z)) + 12*i*sqrt(3)*z^(1/4))^(1/3) + (4*sqrt(4 - 27*sqrt(z)) - 12*i*sqrt(3)*z^(1/4))^(1/3))/(4*sqrt(4 - 27*sqrt(z))), and

A2 = (1/(4*sqrt(4 + 27*sqrt(z)) + 12*sqrt(3)*z^(1/4))^(1/3) + 1/(4*sqrt(4 + 27*sqrt(z)) - 12*sqrt(3)*z^(1/4))^(1/3))/sqrt(4 + 27*sqrt(z)),

where i = sqrt(-1), the imaginary unit. (End)

G.f.: A(x) = 1 + (x*B(x)')/(B(x)), B(x) = (1 + x*B(x)^5)*C(x*B(x)^5) is g.f. of A060941, C(x) is g.f. of Catalan numbers.

a(n) = n*Sum_{i = 0..n}((C(5*n,i)*C(7*n-2*i-1,n-i))/(6*n-i)), n > 1, a(0) = 1.

a(n) = 1/5*A001450(n) for n >= 1. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. - Peter Bala, Oct 05 2015

D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

O.g.f.: 12*hypergeom([3/5, 4/5, 1, 6/5, 7/5], [2/3, 4/3, 3/2, 2], (3125*x)/108).

E.g.f.: 12*hypergeom([3/5, 4/5, 6/5, 7/5], [2/3, 4/3, 3/2, 2], (3125*x)/108).

O.g.f. denoted by h(x), satisfies the algebraic equation of order 10:

1889568 - 6141096*x + 10628820*x^2 - 59049*x^3 + (-2834352*x^3 + 4861701*x^2 - 2834352*x - 157464)*h(x) + 13122*x*(14*x^3 - 77*x^2 + 124*x + 30)*h(x)^2 - 4374*x^2*(14*x^2 + 94*x + 99)*h(x)^3 + 729*x^3*(50*x^2 + 32*x + 377)*h(x)^4 - 243*x^4*(11*x^2 - 40*x + 456)*h(x)^5 - 243*x^5*(8*x - 121)*h(x)^6 + 54*x^6*(2*x - 95)*h(x)^7 + 567*x^7*h(x)^8 - 36*x^8*h(x)^9 + x^9*h(x)^10 = 0.

a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, where W(x) = W1(x)+W2(x)+W3(x)+W4(x), with

W1(x) = (3*sqrt(5)*csc(Pi/5)*sin(Pi/10)*hypergeom([-2/5, 1/10, 4/15, 14/15], [1/5, 2/5, 4/5], (108*x)/3125))/(2*Pi*x^(2/5)),

W2(x) = (6*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([-1/5, 3/10, 7/15, 17/15], [2/5, 3/5, 6/5], (108*x)/3125))/(5*Pi*x^(1/5)),

W3(x) = -(24*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*x^(1/5)*hypergeom([1/5, 7/10, 13/15, 23/15], [4/5, 7/5, 8/5], (108*x)/3125))/(125*Pi), and

W4(x) = -(33*sqrt(5)*csc(Pi/5)*sin(Pi/10)*x^(2/5)*hypergeom([2/5, 9/10, 16/15, 26/15], [6/5, 8/5, 9/5], (108*x)/3125))/(1250*Pi).

This integral representation is unique as it is the solution of the Hausdorff power moment of the function W(x). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 3125/108. Therefore a(n) is a positive definite sequence.