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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A367937 Expansion of e.g.f. exp(4*(exp(x) - 1) + 3*x).

Original entry on oeis.org

1, 7, 53, 431, 3741, 34471, 335621, 3438943, 36954285, 415187415, 4864054165, 59278367247, 749926582717, 9829744447495, 133267495918885, 1865916660838847, 26942271261464525, 400673643394972983, 6129834703935247285, 96368617886967750767, 1555302323744129219293
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 05 2023

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[4 (Exp[x] - 1) + 3 x], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + 4 Sum[Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(exp(4*(exp(x) - 1) + 3*x))) \\ Michel Marcus, Dec 07 2023

Formula

G.f. A(x) satisfies: A(x) = 1 + x * ( 3 * A(x) + 4 * A(x/(1 - x)) / (1 - x) ).
a(n) = exp(-4) * Sum_{k>=0} 4^k * (k+3)^n / k!.
a(0) = 1; a(n) = 3 * a(n-1) + 4 * Sum_{k=1..n} binomial(n-1,k-1) * a(n-k).

A332188 a(n) = (1/e^n) * Sum_{j>=2} j^n * n^j / (j-2)!.

Original entry on oeis.org

0, 3, 72, 1557, 36928, 986550, 29641608, 994006209, 36887753216, 1502798312547, 66730937637400, 3209318261685690, 166242143849148864, 9229638177763268395, 546842961612529341032, 34443269219453881669425
Offset: 0

Views

Author

Pedro Caceres, Oct 30 2020

Keywords

Examples

			a(3) = 1557 = (1/e^3) * Sum_{j>=2} j^3 * 3^j / factorial(j-2).
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[n^k*(StirlingS2[n + 2, k] - StirlingS2[n + 1, k]), {k, 2, n + 2}]; Array[a, 16, 0] (* Amiram Eldar, Oct 30 2020 *)
  • PARI
    a(n) = sum(k=0, n+2, n^k*(stirling(n+2,k,2) - stirling(n+1,k,2))); \\ Michel Marcus, Oct 30 2020
  • SageMath
    # Increase precision for larger n!
    R = RealField(100)
    t = 2
    sol = [0]*18
    for n in range(0, 18):
        suma = R(0)
        for j in range(t, 1000):
            suma += (j^n * n^j) / factorial(j - t)
        suma *= exp(-n)
        sol[n] = round(suma)
    print(sol) # Thanks to Peter Luschny for his example in A338282.
    

Formula

a(n) = Sum_{k=0..n+2} n^k*(Stirling2(n+2,k) - Stirling2(n+1,k)). [Thanks to Andrew Howroyd for his example in A338282]
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