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 1-5 of 5 results.

A337038 a(n) = exp(-1/2) * Sum_{k>=0} (2*k - 1)^n / (2^k * k!).

Original entry on oeis.org

1, 0, 2, 4, 20, 96, 552, 3536, 25104, 194816, 1637408, 14792768, 142761280, 1464117760, 15886137984, 181667507456, 2182268117248, 27456279388160, 360872502280704, 4943580063237120, 70437638474568704, 1041911242274562048, 15972832382065977344, 253388070573020401664
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 12 2020

Keywords

Links

Crossrefs

Cf. A000296, A004211, A007405, A124311, A166922, A217203, A337039, A337040, A337041, A337042, A337043.

Programs

  • Maple
    E:= exp((exp(2*x)-1)/2-x):
S:= series(E,x,31):
seq(coeff(S,x,i)*i!,i=0..30); # Robert Israel, Aug 26 2020
  • Mathematica
    nmax = 23; CoefficientList[Series[Exp[(Exp[2 x] - 1)/2 - x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 2^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 23}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 2^k BellB[k, 1/2], {k, 0, n}], {n, 0, 23}]

Formula

G.f. A(x) satisfies: A(x) = (1 - 2*x + x*A(x/(1 - 2*x))) / (1 - x - 2*x^2).
G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 2*j*x/(1 + x)).
E.g.f.: exp((exp(2*x) - 1) / 2 - x).
a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 2^k * a(n-k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A004211(k).
a(n) ~ 2^(n - 1/2) * n^(n - 1/2) * exp(n/LambertW(2*n) - n - 1/2) / (sqrt(1 + LambertW(2*n)) * LambertW(2*n)^(n - 1/2)). - Vaclav Kotesovec, Jun 26 2022

A337039 a(n) = exp(-1/3) * Sum_{k>=0} (3*k - 1)^n / (3^k * k!).

Original entry on oeis.org

1, 0, 3, 9, 54, 351, 2673, 22842, 216513, 2248965, 25351704, 307699965, 3995419365, 55207193328, 808078734999, 12480510487509, 202697232446070, 3451417004044323, 61450890989472837, 1141331486235356178, 22066085726516137149, 443236553318792110113, 9233934519951699602400
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 12 2020

Keywords

Crossrefs

Cf. A000296, A003575, A004212, A337038, A337040, A337041, A337042, A337043.

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[(Exp[3 x] - 1)/3 - x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 3^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 22}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 3^k BellB[k, 1/3], {k, 0, n}], {n, 0, 22}]

Formula

G.f. A(x) satisfies: A(x) = (1 - 3*x + x*A(x/(1 - 3*x))) / (1 - 2*x - 3*x^2).
G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 3*j*x/(1 + x)).
E.g.f.: exp((exp(3*x) - 1) / 3 - x).
a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 3^k * a(n-k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A004212(k).
a(n) ~ 3^(n - 1/3) * n^(n - 1/3) * exp(n/LambertW(3*n) - n - 1/3) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^(n - 1/3)). - Vaclav Kotesovec, Jun 26 2022

A337040 a(n) = exp(-1/4) * Sum_{k>=0} (4*k - 1)^n / (4^k * k!).

Original entry on oeis.org

1, 0, 4, 16, 112, 896, 8384, 88320, 1032448, 13242368, 184591360, 2773929984, 44641579008, 765196926976, 13905753980928, 266855007453184, 5388980396818432, 114172599765827584, 2530858142594760704, 58556990344729198592, 1411095950792925904896, 35347148031264582270976
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 12 2020

Keywords

Crossrefs

Cf. A000296, A003576, A004213, A337038, A337039, A337041, A337042, A337043.

Programs

  • Mathematica
    nmax = 21; CoefficientList[Series[Exp[(Exp[4 x] - 1)/4 - x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 4^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 21}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 4^k BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

Formula

G.f. A(x) satisfies: A(x) = (1 - 4*x + x*A(x/(1 - 4*x))) / (1 - 3*x - 4*x^2).
G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 4*j*x/(1 + x)).
E.g.f.: exp((exp(4*x) - 1) / 4 - x).
a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 4^k * a(n-k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A004213(k).
a(n) ~ 4^(n - 1/4) * n^(n - 1/4) * exp(n/LambertW(4*n) - n - 1/4) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^(n - 1/4)). - Vaclav Kotesovec, Jun 26 2022

A337041 a(n) = exp(-1/5) * Sum_{k>=0} (5*k - 1)^n / (5^k * k!).

Original entry on oeis.org

1, 0, 5, 25, 200, 1875, 20625, 256250, 3534375, 53515625, 881468750, 15667578125, 298478828125, 6060493750000, 130542772265625, 2971013486328125, 71193375156250000, 1790666151318359375, 47145509926611328125, 1296156682961425781250, 37129279010879638671875
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 12 2020

Keywords

Crossrefs

Cf. A000296, A003577, A005011, A337038, A337039, A337040, A337042, A337043.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[(Exp[5 x] - 1)/5 - x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 5^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 20}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 5^k BellB[k, 1/5], {k, 0, n}], {n, 0, 20}]

Formula

G.f. A(x) satisfies: A(x) = (1 - 5*x + x*A(x/(1 - 5*x))) / (1 - 4*x - 5*x^2).
G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 5*j*x/(1 + x)).
E.g.f.: exp((exp(5*x) - 1) / 5 - x).
a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 5^k * a(n-k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A005011(k).
a(n) ~ 5^(n - 1/5) * n^(n - 1/5) * exp(n/LambertW(5*n) - n - 1/5) / (sqrt(1 + LambertW(5*n)) * LambertW(5*n)^(n - 1/5)). - Vaclav Kotesovec, Jun 26 2022

A337042 a(n) = exp(-1/6) * Sum_{k>=0} (6*k - 1)^n / (6^k * k!).

Original entry on oeis.org

1, 0, 6, 36, 324, 3456, 43416, 618192, 9778320, 169827840, 3210376032, 65540155968, 1435094563392, 33510354739200, 830486180748672, 21756166766173440, 600339119317643520, 17394883290643709952, 527782830161632077312, 16727350847049194775552
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 12 2020

Keywords

Comments

In general, if m >= 1, b <> 0 and e.g.f. = exp(m*exp(b*x) + r*x + s) then a(n) ~ b^n * n^(n + r/b) * exp(n/LambertW(n/m) - n + s) / (m^(r/b) * sqrt(1 + LambertW(n/m)) * LambertW(n/m)^(n + r/b)). - Vaclav Kotesovec, Jun 28 2022

Crossrefs

Cf. A000296, A003578, A005012, A337038, A337039, A337040, A337041, A337043.

Programs

  • Mathematica
    nmax = 19; CoefficientList[Series[Exp[(Exp[6 x] - 1)/6 - x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 6^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 19}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 6^k BellB[k, 1/6], {k, 0, n}], {n, 0, 19}]

Formula

G.f. A(x) satisfies: A(x) = (1 - 6*x + x*A(x/(1 - 6*x))) / (1 - 5*x - 6*x^2).
G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 6*j*x/(1 + x)).
E.g.f.: exp((exp(6*x) - 1) / 6 - x).
a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 6^k * a(n-k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A005012(k).
a(n) ~ 6^(n - 1/6) * n^(n - 1/6) * exp(n/LambertW(6*n) - n - 1/6) / (sqrt(1 + LambertW(6*n)) * LambertW(6*n)^(n - 1/6)). - Vaclav Kotesovec, Jun 26 2022
Showing 1-5 of 5 results.

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