A295765 -id:A295765 - cp's OEIS frontend

A143669 a(n) = binomial((n+1)^2, n) / (n+1)^2.

1, 1, 4, 35, 506, 10472, 285384, 9706503, 397089550, 19022318084, 1045659267016, 64924369564353, 4496010926381352, 343688726144945040, 28753733905585301136, 2613784129155164386575, 256569498889138342791510, 27050758656206146528056236

Offset: 0

Paul D. Hanna, Aug 28 2008

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 + 1)). Cf. A135862.
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.

Cf. A295765, A135862, A182316.

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

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

PARI
```
a(n)=binomial((n+1)^2,n)/(n+1)^2
```

A295763 G.f. satisfies: A(x) = Sum_{n>=0} binomial(n(n+1),n)/(n+1) x^n/A(x)^n.

Original entry on oeis.org

1, 1, 4, 42, 744, 18570, 596929, 23457763, 1089601420, 58424516424, 3553205095552, 241765128267597, 18202737707568180, 1502857964050898494, 135033771405912550765, 13119213786776385900734, 1370572549521961522812200, 153224265349198540163190599, 18253426026439076436840194131, 2308479498698233016622014842489

Offset: 0

Paul D. Hanna, Jan 06 2018

nonn

			G.f.: A(x) = 1 + x + 4*x^2 + 42*x^3 + 744*x^4 + 18570*x^5 + 596929*x^6 + 23457763*x^7 + 1089601420*x^8 + 58424516424*x^9 + 3553205095552*x^10 +...
such that
A(x) = 1 + x/A(x) + 5*(x/A(x))^2 + 55*(x/A(x))^3 + 969*(x/A(x))^4 + 23751*(x/A(x))^5 + 749398*(x/A(x))^6 +...+ binomial(n*(n+1),n)/(n+1)*(x/A(x))^n +...
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 1, 4, 42, 744, 18570, 596929, 23457763, 1089601420, ...];
  [1, 2, 9, 92, 1588, 38964, 1238714, 48320440, 2233007214, ...];
  [1, 3, 15, 151, 2544, 61356, 1928659, 74668905, 3432698217, ...];
  [1, 4, 22, 220, 3625, 85936, 2670332, 102589280, 4691284160, ...];
  [1, 5, 30, 300, 4845, 112911, 3467585, 132173305, 6011511390, ...];
  [1, 6, 39, 392, 6219, 142506, 4324575, 163518732, 7396271082, ...];
  [1, 7, 49, 497, 7763, 174965, 5245786, 196729744, 8848607971, ...];
  [1, 8, 60, 616, 9494, 210552, 6236052, 231917400, 10371729633, ...];
  [1, 9, 72, 750, 11430, 249552, 7300581, 269200107, 11969016345, ...]; ...
in which the main diagonal begins:
  [1, 2, 15, 220, 4845, 142506, 5245786, ..., binomial(n*(n+1),n), ...],
thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^(n*(n+1)) for n>=0.

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

cf. A295764, A295765, A135861, A135860.

terms = 20; A[] = 1; Do[A[x] = Sum[Binomial[n*(n+1), n]/(n+1)*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*(m+1),m) - 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*(n+1),n) for n>=0.

a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(1/2 - exp(-2)) / sqrt(2*Pi) = 0.5744892944370457395619... - Vaclav Kotesovec, Oct 17 2020, updated Mar 18 2024

A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

Original entry on oeis.org

1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, 130495143060, 8002598818976, 548083639109204, 41481408076217480, 3439242379130298624, 310088460754461374608, 30213233395705566500248, 3164041937175571224372118, 354458174755809132668847604, 42301583754989091491716706272, 5358139267082652649000120167456

Offset: 0

Paul D. Hanna, Jan 06 2018

nonn

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 84*x^3 + 1522*x^4 + 38940*x^5 + 1278880*x^6 + 51136512*x^7 + 2407921070*x^8 + 130495143060*x^9 + 8002598818976*x^10 +...
such that
A(x) = 1 + 2*(x/A(x)) + 12*(x/A(x))^2 + 140*(x/A(x))^3 + 2530*(x/A(x))^4 + 62832*(x/A(x))^5 + 1997688*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)*(x/A(x))^n +...
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, ...];
  [1, 4, 20, 200, 3444, 85312, 2744928, 108267280, 5049708672, ...];
  [1, 6, 36, 356, 5862, 140508, 4424984, 172064160, 7946443350, ...];
  [1, 8, 56, 560, 8888, 206176, 6350112, 243284064, 11121338640, ...];
  [1, 10, 80, 820, 12650, 284252, 8556240, 322780800, 14599990830, ...];
  [1, 12, 108, 1144, 17292, 376992, 11084864, 411518448, 18410660208, ...];
  [1, 14, 140, 1540, 22974, 487004, 13983816, 510586400, 22584587382, ...];
  [1, 16, 176, 2016, 29872, 617280, 17308096, 621216192, 27156348512, ...];
  [1, 18, 216, 2580, 38178, 771228, 21120768, 744800256, 32164253550, ...]; ...
in which the main diagonal begins:
  [1, 4, 36, 560, 12650, 376992, 13983816, ..., binomial((n+1)^2,n), ...],
thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^((n+1)^2) for n>=0.

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

Cf. A295765, A295763, A070780.

terms = 20; A[] = 1; Do[A[x] = Sum[Binomial[(n+1)^2, n]/(n+1)*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) - 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) for n>=0.

a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(3/2 - 2*exp(-2)) / sqrt(2*Pi) = 1.3639581816297942453935... - Vaclav Kotesovec, Aug 11 2021, updated Apr 20 2024

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A143669 a(n) = binomial((n+1)^2, n) / (n+1)^2.

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A295763 G.f. satisfies: A(x) = Sum_{n>=0} binomial(n(n+1),n)/(n+1) x^n/A(x)^n.

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A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

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cp's OEIS Frontend

A143669 a(n) = binomial((n+1)^2, n) / (n+1)^2.

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Magma

Mathematica

PARI

A295763 G.f. satisfies: A(x) = Sum_{n>=0} binomial(n*(n+1),n)/(n+1) * x^n/A(x)^n.

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Formula

A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

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Formula

A295763 G.f. satisfies: A(x) = Sum_{n>=0} binomial(n(n+1),n)/(n+1) x^n/A(x)^n.