A135862 a(n) = binomial(n*(n+1),n)/(n^2+1).

1, 1, 3, 22, 285, 5481, 141778, 4638348, 184138713, 8612835715, 464333035881, 28368973183200, 1938034271677595, 146439782923866810, 12129248258088578100, 1092975597259714940696, 106463932364272178140209, 11148388021809242372111895, 1248977393391271550436645787

Offset: 0

Author

Paul D. Hanna, Dec 02 2007

Keywords

• nonn
• easy

Comments

From Peter Bala, Dec 02 2015: (Start)

Let x = p/q be a positive rational in reduced form with p,q > 0. Define Cat(x) = 1/(2*p + q)*binomial(2*p + q, p). Then Cat(n) = Catalan(n). This sequence is Cat(n/(n^2 - n + 1)). Cf. A143669.

See Armstrong et al. for combinatorial interpretations of these generalized Catalan number sequences. (End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..339
• D. Armstrong, B. Rhoades, and N. Williams, Rational associahedra and noncrossing partitions arxiv:1305.7286v1 [math.CO], 2013.
• R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A135860, A135861, A143669.

Programs

Magma

[Binomial(n*(n+1),n)/(n^2+1): n in [0..20]]; // Vincenzo Librandi, Dec 09 2015

Mathematica

Table[Binomial[n (n + 1), n]/(n^2 + 1), {n, 0, 30}] (* Vincenzo Librandi, Dec 09 2015 *)

PARI

a(n)=binomial(n*(n+1),n)/(n^2+1)
for(n=0,30,print1(a(n),", "))

PARI

{a(n) = local(A=1+x); for(i=1, n, A = exp( sum(k=1, n, A^(n*k)*x^k/k +x*O(x^n)))); polcoeff(A, n)}
for(n=0,30,print1(a(n),", "))

Formula

a(n) = A135860(n)/(n^2+1).

a(n) = 1/(n^2 + n + 1)*binomial(n^2 + n + 1, n). - Peter Bala, Dec 02 2015

Ccongruences: a(p) == (1 + p - p^2 - p^3) ( mod p^4 ) for prime p >= 5 and a(2*p) == (1 + 6*p + 4*p^2) ( mod p^3 ) for all prime p (apply Mestrovic, equation 37). - Peter Bala, Feb 23 2020