A143669 -id:A143669 - cp's OEIS frontend

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

Paul D. Hanna, Dec 02 2007

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)

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.

Cf. A135860, A135861, A143669.

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

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

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

{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),", "))

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

A295765 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n/A(x)^n.

Original entry on oeis.org

1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, 15487290535, 859400072837, 53749578759526, 3743585586509849, 287496351622105328, 24143937833744911767, 2201703647718624364913, 216700738558116024114289, 22900073562659910815354339, 2586409916780162599516986945, 310947096149155992699450689912, 39650252031533561961437812566315

Offset: 0

Paul D. Hanna, Jan 06 2018

nonn

			G.f.: A(x) = 1 + x + 3*x^2 + 25*x^3 + 369*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + 859400072837*x^10 + ...
such that
A(x) = 1 + x/A(x) + 4*(x/A(x))^2 + 35*(x/A(x))^3 + 506*(x/A(x))^4 + 10472*(x/A(x))^5 + 285384*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)^2*(x/A(x))^n + ...
RELATED SERIES.
Define B(x) = A(x*B(x)) and A(x) = B(x/A(x)) then B(x) begins
B(x) = 1 + x + 4*x^2 + 35*x^3 + 506*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + ... + binomial((n+1)^2,n)/(n+1)^2*x^n + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, ...];
  [1, 2, 7, 56, 797, 16650, 460291, 15862152, 655825337, ...];
  [1, 3, 12, 94, 1293, 26409, 719922, 24587202, 1010428347, ...];
  [1, 4, 18, 140, 1867, 37272, 1001476, 33887832, 1384043365, ...];
  [1, 5, 25, 195, 2530, 49366, 1306860, 43802060, 1777652015, ...];
  [1, 6, 33, 260, 3294, 62832, 1638166, 54370836, 2192294775, ...];
  [1, 7, 42, 336, 4172, 77826, 1997688, 65638294, 2629075183, ...];
  [1, 8, 52, 424, 5178, 94520, 2387940, 77652024, 3089164371, ...];
  [1, 9, 63, 525, 6327, 113103, 2811675, 90463365, 3573805950, ...]; ...
in which the main diagonal begins:
  [1, 2, 12, 140, 2530, 62832, 1997688, ..., binomial((n+1)^2,n)/(n+1), ...].

Paul D. Hanna, Table of n, a(n) for n = 0..300

cf. A295764, A295763, A143669.

terms = 21; A[] = 1; Do[A[x] = Sum[Binomial[(n+1)^2, n]/(n+1)^2*x^n/ A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)

{a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m)/(m+1) - V[m+1])/(m+1);); A[n+1]}
for(n=0,20,print1(a(n),", "))

G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n)/(n+1) for n>=0.

a(n) ~ c * exp(n) * n^(n - 5/2), where c = exp(3/2 - exp(-2)) / sqrt(2*Pi) = 1.56162380971247949723297... - Vaclav Kotesovec, Oct 17 2020, updated Apr 20 2024

A182316 a(n) = binomial(n^2 + 3*n, n) / (n+1)^2.

Original entry on oeis.org

1, 1, 5, 51, 819, 18278, 527085, 18730855, 793542167, 39113958819, 2201663313200, 139461523272085, 9824294829146550, 762188806010669820, 64595315110014533629, 5939055918736259991759, 588894813538193130767295, 62651281502108852275337225

Offset: 0

Paul D. Hanna, Apr 24 2012

a(n) = < PF_n, PF_n >, where PF_n is the parking function symmetric function and <,> denotes the usual scalar product on symmetric functions (proved). - Richard Stanley, Sep 24 2015

Seiichi Manyama, Table of n, a(n) for n = 0..339

Cf. A143669.

A182316:=n->binomial(n^2 + 3*n, n) / (n+1)^2: seq(A182316(n), n=0..20); # Wesley Ivan Hurt, Feb 11 2017

{a(n)=binomial((n+1)^2+n-1, n)/(n+1)^2}
for(n=0,20,print1(a(n),","))

a(n) = [x^n] 1/(1-x)^((n+1)^2) / (n+1)^2 ; that is, a(n) equals the coefficient of x^n in 1/(1-x)^((n+1)^2) divided by (n+1)^2.

A299434 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n / A(x)^((n+1)^2).

Original entry on oeis.org

1, 1, 1, 6, 77, 1451, 35730, 1082481, 38913817, 1619979291, 76724619427, 4077896446598, 240566693095072, 15609120639706252, 1105414601508493001, 84881459931003622118, 7026832554316541379141, 624014794413319426058889, 59184228450018585954486975, 5971678912361406721742217080, 638782082648832471805820934833, 72213308562202419209594988387550, 8603323896642095980014195130664418

Offset: 0

Paul D. Hanna, Feb 13 2018

nonn

Compare to: 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.

			G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 77*x^4 + 1451*x^5 + 35730*x^6 + 1082481*x^7 + 38913817*x^8 + 1619979291*x^9 + 76724619427*x^10 +...
such that
1 = 1/A(x) + C(4,1)/4*x/A(x)^4 + C(9,2)/9*x^2/A(x)^9 + C(16,3)/16*x^3/A(x)^16 + C(25,4)/25*x^4/A(x)^25 + C(36,5)/36*x^5/A(x)^36 + C(49,6)/49*x^6/A(x)^49 + ...
more explicitly,
1 = 1/A(x) + x/A(x)^4 + 4*x^2/A(x)^9 + 35*x^3/A(x)^16 + 506*x^4/A(x)^25 + 10472*x^5/A(x)^36 + 285384*x^6/A(x)^49 + ... + A143669(n)*x^n/A(x)^((n+1)^2) + ...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A298692, A143669.

{a(n) = my(A=[1]); for(i=1,n, A = Vec(sum(n=0,#A,binomial((n+1)^2,n)/(n+1)^2 * x^n/Ser(A)^((n+1)^2-1) )));G=Ser(A);A[n+1]}
for(n=0,30,print1(a(n),", "))

A227469 a(n) = binomial((n+1)^2, n) * (2*n+1) / (n+1)^2 for n>=0.

Original entry on oeis.org

1, 3, 20, 245, 4554, 115192, 3709992, 145597545, 6750522350, 361424043596, 21958844607336, 1493260499980119, 112400273159533800, 9279595605913516080, 833858283261973732944, 81027308003810095983825, 8466793463341565312119830, 946776552967215128481968260, 112816032067073649248569919400

Offset: 0

Paul D. Hanna, Aug 23 2013

nonn

			G.f.: A(x) = 1 + 3*x + 20*x^2 + 245*x^3 + 4554*x^4 + 115192*x^5 +...

Cf. A143669.

{a(n)=binomial((n+1)^2,n)*(2*n+1)/(n+1)^2}
for(n=0,20,print1(a(n),", "))

a(n) = (2*n+1)*A143669(n).

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

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A295765 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n/A(x)^n.

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A182316 a(n) = binomial(n^2 + 3*n, n) / (n+1)^2.

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A299434 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n / A(x)^((n+1)^2).

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A227469 a(n) = binomial((n+1)^2, n) * (2*n+1) / (n+1)^2 for n>=0.

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