cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

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

Views

Author

Paul D. Hanna, Jan 06 2018

Keywords

Examples

			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.

Links

Crossrefs

cf. A295764, A295765, A135861, A135860.

Programs

  • Mathematica
    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 *)
  • PARI
    {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),", "))

Formula

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

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

Views

Author

Paul D. Hanna, Jan 06 2018

Keywords

Examples

			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), ...].

Links

Crossrefs

cf. A295764, A295763, A143669.

Programs

  • Mathematica
    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 *)
  • PARI
    {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),", "))

Formula

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
Showing 1-2 of 2 results.

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