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 1-7 of 7 results.

A347002 Expansion of e.g.f. exp( -log(1 - x)^3 / 3! ).

Original entry on oeis.org

1, 0, 0, 1, 6, 35, 235, 1834, 16352, 163764, 1818030, 22143726, 293476326, 4203311892, 64682865156, 1064154324024, 18636296872320, 346103784493560, 6793394350116600, 140508244952179200, 3054120126193160280, 69596730438090806880, 1659041650323705102840
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[-Log[1 - x]^3/3!], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] Abs[StirlingS1[k, 3]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
  • PARI
    a(n) = sum(k=0, n\3, (3*k)!*abs(stirling(n, 3*k, 1))/(6^k*k!)); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * |Stirling1(k,3)| * a(n-k).
a(n) = Sum_{k=0..floor(n/3)} (3*k)! * |Stirling1(n,3*k)|/(6^k * k!). - Seiichi Manyama, May 06 2022

A346924 Expansion of e.g.f. 1 / (1 + log(1 - x)^5 / 5!).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 15, 175, 1960, 22449, 269577, 3430790, 46480830, 671260876, 10329270952, 169125055736, 2940784282800, 54182845939104, 1055291277366108, 21674715826211532, 468366193441002564, 10624074081842024496, 252432685158931968768, 6270222495850552958004
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 07 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[1/(1 + Log[1 - x]^5/5!), {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] Abs[StirlingS1[k, 5]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(1/(1+log(1-x)^5/5!))) \\ Michel Marcus, Aug 07 2021
    
  • PARI
    a(n) = sum(k=0, n\5, (5*k)!*abs(stirling(n, 5*k, 1))/120^k); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * |Stirling1(k,5)| * a(n-k).
a(n) ~ n! * 2^(3/5) * 3^(1/5) * exp(2^(3/5)*15^(1/5)*n) / (5^(4/5) * (exp(2^(3/5)*15^(1/5)) - 1)^(n+1)). - Vaclav Kotesovec, Aug 08 2021
a(n) = Sum_{k=0..floor(n/5)} (5*k)! * |Stirling1(n,5*k)|/120^k. - Seiichi Manyama, May 06 2022

A347001 Expansion of e.g.f. exp( log(1 - x)^2 / 2 ).

Original entry on oeis.org

1, 0, 1, 3, 14, 80, 544, 4284, 38310, 383256, 4239006, 51345690, 675770028, 9600349824, 146396925648, 2384700728760, 41320373582652, 758780222426592, 14718569154071964, 300706641183038292, 6453691377726073128, 145154958710291611200, 3414131149418742544320
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[Log[1 - x]^2/2], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] Abs[StirlingS1[k, 2]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
  • PARI
    a(n) = sum(k=0, n\2, (2*k)!*abs(stirling(n, 2*k, 1))/(2^k*k!)); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * |Stirling1(k,2)| * a(n-k).
a(n) = Sum_{k=0..floor(n/2)} (2*k)! * |Stirling1(n,2*k)|/(2^k * k!). - Seiichi Manyama, May 06 2022

A347003 Expansion of e.g.f. exp( log(1 - x)^4 / 4! ).

Original entry on oeis.org

1, 0, 0, 0, 1, 10, 85, 735, 6804, 68544, 754130, 9044750, 117773656, 1656897528, 25061576176, 405667844400, 6997383182854, 128126051451184, 2481884332498848, 50702417505257904, 1089371806098805286, 24555007848629510700, 579348221233739760550, 14278529041496660104450
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[Exp[Log[1 - x]^4/4!], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] Abs[StirlingS1[k, 4]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]
  • PARI
    a(n) = sum(k=0, n\4, (4*k)!*abs(stirling(n, 4*k, 1))/(24^k*k!)); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * |Stirling1(k,4)| * a(n-k).
a(n) = Sum_{k=0..floor(n/4)} (4*k)! * |Stirling1(n,4*k)|/(24^k * k!). - Seiichi Manyama, May 06 2022

A353404 Expansion of e.g.f. exp(-log(1 - x)^5).

Original entry on oeis.org

1, 0, 0, 0, 0, 120, 1800, 21000, 235200, 2693880, 34133400, 509823600, 9012207600, 180053908320, 3870208261920, 87083930169600, 2034907999488000, 49491370609706880, 1259740748821328640, 33710658096392887680, 949893399326820528000
Offset: 0

Views

Author

Seiichi Manyama, May 06 2022

Keywords

Crossrefs

Column k=5 of A357882.

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Exp[-Log[1-x]^5],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Mar 13 2023 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(-log(1-x)^5)))
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace((1-x)^(-log(1-x)^4)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=120*sum(j=1, i, binomial(i-1, j-1)*abs(stirling(j, 5, 1))*v[i-j+1])); v;
    
  • PARI
    a(n) = sum(k=0, n\5, (5*k)!*abs(stirling(n, 5*k, 1))/k!);

Formula

E.g.f.: (1 - x)^(-(log(1 - x))^4).
a(0) = 1; a(n) = 120 * Sum_{k=1..n} binomial(n-1,k-1) * |Stirling1(k,5)| * a(n-k).
a(n) = Sum_{k=0..floor(n/5)} (5*k)! * |Stirling1(n,5*k)|/k!.

A354137 Expansion of e.g.f. exp(log(1 + x)^5/120).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, -15, 175, -1960, 22449, -269199, 3410000, -45753180, 650179816, -9771920158, 155020385156, -2589888417480, 45461879164584, -836540418765834, 16099972965770778, -323385447259166454, 6764948641797695496, -147088325599708573080
Offset: 0

Views

Author

Seiichi Manyama, May 18 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(log(1+x)^5/120)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, binomial(i-1, j-1)*stirling(j, 5, 1)*v[i-j+1])); v;
    
  • PARI
    a(n) = sum(k=0, n\5, (5*k)!*stirling(n, 5*k, 1)/(120^k*k!));

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * Stirling1(k,5) * a(n-k).
a(n) = Sum_{k=0..floor(n/5)} (5*k)! * Stirling1(n,5*k)/(120^k * k!).

A357883 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} (k*j)!* |Stirling1(n,k*j)|/(k!^j * j!).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 6, 0, 1, 0, 0, 3, 24, 0, 1, 0, 0, 1, 14, 120, 0, 1, 0, 0, 0, 6, 80, 720, 0, 1, 0, 0, 0, 1, 35, 544, 5040, 0, 1, 0, 0, 0, 0, 10, 235, 4284, 40320, 0, 1, 0, 0, 0, 0, 1, 85, 1834, 38310, 362880, 0, 1, 0, 0, 0, 0, 0, 15, 735, 16352, 383256, 3628800, 0
Offset: 0

Views

Author

Seiichi Manyama, Oct 18 2022

Keywords

Examples

			Square array begins:
  1,   1,  1,  1,  1, 1, ...
  0,   1,  0,  0,  0, 0, ...
  0,   2,  1,  0,  0, 0, ...
  0,   6,  3,  1,  0, 0, ...
  0,  24, 14,  6,  1, 0, ...
  0, 120, 80, 35, 10, 1, ...
		

Crossrefs

Programs

  • PARI
    T(n, k) = sum(j=0, n, (k*j)!*abs(stirling(n, k*j, 1))/(k!^j*j!));
    
  • PARI
    T(n, k) = if(k==0, 0^n, n!*polcoef(exp((-log(1-x+x*O(x^n)))^k/k!), n));

Formula

For k > 0, e.g.f. of column k: exp((-log(1-x))^k / k!).
T(0,k) = 1; T(n,k) = Sum_{j=1..n} binomial(n-1,j-1) * |Stirling1(j,k)| * T(n-j,k).
Showing 1-7 of 7 results.