A135862 - OEIS

%I #30 Sep 08 2022 08:45:32

%S 1,1,3,22,285,5481,141778,4638348,184138713,8612835715,464333035881,

%T 28368973183200,1938034271677595,146439782923866810,

%U 12129248258088578100,1092975597259714940696,106463932364272178140209,11148388021809242372111895,1248977393391271550436645787

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

%C From _Peter Bala_, Dec 02 2015: (Start)

%C 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.

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

%H Seiichi Manyama, <a href="/A135862/b135862.txt">Table of n, a(n) for n = 0..339</a>

%H D. Armstrong, B. Rhoades, and N. Williams, <a href="http://arxiv.org/abs/1305.7286">Rational associahedra and noncrossing partitions</a> arxiv:1305.7286v1 [math.CO], 2013.

%H R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

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

%F a(n) = 1/(n^2 + n + 1)*binomial(n^2 + n + 1, n). - _Peter Bala_, Dec 02 2015

%F 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

%t Table[Binomial[n (n + 1), n]/(n^2 + 1), {n, 0, 30}] (* _Vincenzo Librandi_, Dec 09 2015 *)

%o (PARI) a(n)=binomial(n*(n+1),n)/(n^2+1)

%o for(n=0,30,print1(a(n),", "))

%o (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)}

%o for(n=0,30,print1(a(n),", "))

%o (Magma) [Binomial(n*(n+1),n)/(n^2+1): n in [0..20]]; // _Vincenzo Librandi_, Dec 09 2015

%Y Cf. A135860, A135861, A143669.

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Dec 02 2007