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.

Showing 1-3 of 3 results.

A355291 Expansion of e.g.f. exp(exp(x)*(exp(x) + 1) - 2).

Original entry on oeis.org

1, 3, 14, 81, 551, 4266, 36803, 348543, 3583484, 39652659, 468970211, 5894584812, 78366374813, 1097537989671, 16136598952718, 248309032411485, 3988468487017379, 66715970326561170, 1159712730763363991, 20909709414253764819, 390374806223071148084, 7534929383736826736007
Offset: 0

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Author

Vaclav Kotesovec, Jun 27 2022

Keywords

Comments

In general, if m > 0, b > d >= 1 and e.g.f. = exp(m*exp(b*x) + r*exp(d*x) + s) then a(n) ~ exp(m*exp(b*z) + r*exp(d*z) + s - n) * (n/z)^(n + 1/2) / sqrt(m*b*(1 + b*z)*exp(b*z) + r*d*(1 + d*z)*exp(d*z)), where z = LambertW(n/m)/b - 1/(d + b/LambertW(n/m) + b^2 * m^(d/b) * n^(1 - d/b) * (1 + LambertW(n/m)) / (d*r*LambertW(n/m)^(2 - d/b))). - Vaclav Kotesovec, Jul 03 2022
In addition, if b/d >=2 then a(n) ~ c * (b*n/LambertW(n/m))^n * exp(n/LambertW(n/m) + r * (n/(m*LambertW(n/m)))^(d/b) - n + s) / sqrt(1 + LambertW(n/m)), where c = 1 for b/d > 2 and c = exp(-r^2/(8*m)) for b/d = 2. - Vaclav Kotesovec, Jul 10 2022

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[2*x] - 2 + Exp[x]], {x, 0, nmax}], x] * Range[0, nmax]!
    Table[Sum[Binomial[n, k] * 2^k * BellB[k] * BellB[n-k], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(exp(exp(x)*(exp(x) + 1) - 2))) \\ Michel Marcus, Jun 27 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 2^k * Bell(k) * Bell(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} (1 + 2^k) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jul 01 2022
a(n) ~ exp(exp(2*z) + exp(z) - 2 - n) * (n/z)^(n + 1/2) / sqrt(2*(1 + 2*z)*exp(2*z) + (1 + z)*exp(z)), where z = LambertW(n)/2 - 1/(1 + 2/LambertW(n) + 4 * n^(1/2) * (1 + LambertW(n)) / LambertW(n)^(3/2)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ 2^n * n^n / (sqrt(1 + LambertW(n)) * LambertW(n)^n * exp(n + 17/8 - n/LambertW(n) - sqrt(n/LambertW(n)))). - Vaclav Kotesovec, Jul 08 2022

A355378 Expansion of e.g.f. exp(exp(3*x) - exp(x)).

Original entry on oeis.org

1, 2, 12, 82, 688, 6754, 75096, 928386, 12591392, 185384130, 2938319144, 49799613538, 897495547184, 17118975292514, 344206910941624, 7270287035936706, 160826794265399360, 3716047107259486082, 89472755268582494792, 2240097688067896960674, 58207872357772581544272
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 30 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Exp[3*x] - Exp[x]], {x, 0, nmax}], x] * Range[0, nmax]!
    Table[Sum[Binomial[n,k] * 3^k * BellB[k] * BellB[n-k, -1], {k, 0, n}], {n, 0, 20}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(3*x) - exp(x)))) \\ Michel Marcus, Jun 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * 3^k * Bell(k) * Bell(n-k, -1).
a(0) = 1; a(n) = Sum_{k=1..n} (3^k - 1) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jun 30 2022
a(n) ~ exp(exp(3*z) - exp(z) - n) * (n/z)^(n + 1/2) / sqrt(3*(1 + 3*z)*exp(3*z) - (1 + z)*exp(z)), where z = LambertW(n)/3 - 1/(1 + 3/LambertW(n) - 9 * n^(2/3) * (1 + LambertW(n)) / LambertW(n)^(5/3)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ (3*n/LambertW(n))^n * exp(n/LambertW(n) - (n/LambertW(n))^(1/3) - n) / sqrt(1 + LambertW(n)). - Vaclav Kotesovec, Jul 10 2022

A355421 Expansion of e.g.f. exp(Sum_{k=1..3} (exp(k*x) - 1)).

Original entry on oeis.org

1, 6, 50, 504, 5870, 76872, 1111646, 17522664, 298133054, 5433157512, 105396184478, 2165189912040, 46901678992958, 1067332196912136, 25435754924426270, 633014456504059368, 16411191933603611198, 442258823578968351624
Offset: 0

Views

Author

Seiichi Manyama, Jul 01 2022

Keywords

Crossrefs

Column k=3 of A355423.

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(sum(k=1, 3, exp(k*x)-1))))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, (1+2^j+3^j)*binomial(i-1, j-1)*v[i-j+1])); v;

Formula

a(0) = 1; a(n) = Sum_{k=1..n} (1 + 2^k + 3^k) * binomial(n-1,k-1) * a(n-k).
Showing 1-3 of 3 results.