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 1-6 of 6 results.

A294289 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f.: exp(Product_{j=1..n} 1/(1+x^j) - 1).

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -1, 3, 0, 1, -1, 1, -13, 0, 1, -1, 1, -1, 73, 0, 1, -1, 1, -7, 25, -501, 0, 1, -1, 1, -7, 73, -241, 4051, 0, 1, -1, 1, -7, 49, -421, 1081, -37633, 0, 1, -1, 1, -7, 49, -181, 2641, -3361, 394353, 0, 1, -1, 1, -7, 49, -301, 1561, -32131, 68881
Offset: 0

Views

Author

Seiichi Manyama, Oct 27 2017

Keywords

Examples

			Square array A(n,k) begins:
   1,    1,    1,    1,    1, ...
   0,   -1,   -1,   -1,   -1, ...
   0,    3,    1,    1,    1, ...
   0,  -13,   -1,   -7,   -7, ...
   0,   73,   25,   73,   49, ...
   0, -501, -241, -421, -181, ...
		

Crossrefs

Columns k=0..5 give A000007, A293125, A294290, A294291, A294292, A294293.
Rows n=0 gives A000012.
Main diagonal gives A294261.
Cf. A294212.

Formula

B(j,k) is the coefficient of Product_{i=1..k} 1/(1+x^i).
A(0,k) = 1 and A(n,k) = (n-1)! * Sum_{j=1..n} j*B(j,k)*A(n-j,k)/(n-j)! for n > 0.

A294250 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f.: exp(Product_{j=1..n} (1+x^j) - 1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 3, 1, 0, 1, 1, 3, 13, 1, 0, 1, 1, 3, 19, 49, 1, 0, 1, 1, 3, 19, 97, 261, 1, 0, 1, 1, 3, 19, 121, 681, 1531, 1, 0, 1, 1, 3, 19, 121, 921, 5971, 9073, 1, 0, 1, 1, 3, 19, 121, 1041, 8491, 50443, 63393, 1, 0, 1, 1, 3, 19, 121, 1041
Offset: 0

Views

Author

Seiichi Manyama, Oct 26 2017

Keywords

Examples

			Square array A(n,k) begins:
   1, 1,   1,   1,   1, ...
   0, 1,   1,   1,   1, ...
   0, 1,   3,   3,   3, ...
   0, 1,  13,  19,  19, ...
   0, 1,  49,  97, 121, ...
   0, 1, 261, 681, 921, ...
		

Crossrefs

Columns k=0..5 give A000007, A000012, A118589, A294251, A294252, A294253.
Rows n=0 gives A000012.
Main diagonal gives A293840.

Formula

B(j,k) is the coefficient of Product_{i=1..k} (1+x^i).
A(0,k) = 1 and A(n,k) = (n-1)! * Sum_{j=1..min(A000217(k),n)} j*B(j,k)*A(n-j,k)/(n-j)! for n > 0.

A294214 E.g.f.: exp(1/((1-x)*(1-x^2)*(1-x^3)) - 1).

Original entry on oeis.org

1, 1, 5, 31, 241, 2261, 25501, 319915, 4564001, 71905321, 1240694101, 23250921431, 470598127825, 10200501671101, 236040247113581, 5800885227542371, 150850086300786241, 4137020164029442385, 119309846230265324581, 3608164806033723494671
Offset: 0

Views

Author

Seiichi Manyama, Oct 25 2017

Keywords

Crossrefs

Column k=3 of A294212.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[E^(1/((1-x)*(1-x^2)*(1-x^3)) - 1), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Dec 02 2021 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2)*(1-x^3))-1)))

Formula

a(n) = a(n-1) + 2*(n-1)*n*a(n-2) + (n-2)*(n-1)*(2*n+1)*a(n-3) - (n-10)*(n-3)*(n-2)*(n-1)*a(n-4) - 4*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5) - (n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-6) + 2*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-7) + 2*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-8) - (n-10)*(n-9)*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-10). - Vaclav Kotesovec, Dec 02 2021
a(n) ~ exp(-101/144 + 29*n^(1/4)/(36*2^(3/4)) + sqrt(n/2) + 2^(7/4)*n^(3/4)/3 - n) * n^(n - 1/8) / 2^(9/8) * (1 + 71323/(103680*2^(1/4)*n^(1/4))). - Vaclav Kotesovec, Dec 02 2021

A294215 E.g.f.: exp(1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) - 1).

Original entry on oeis.org

1, 1, 5, 31, 265, 2501, 29461, 383755, 5721521, 93393865, 1683745381, 32835673751, 693498302905, 15671281854541, 378500195728565, 9704429057721091, 263513260349418721, 7544370749942882705, 227236831102901587141, 7177550671651275241615
Offset: 0

Views

Author

Seiichi Manyama, Oct 25 2017

Keywords

Crossrefs

Column k=4 of A294212.

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Exp[1/((1-x)(1-x^2)(1-x^3)(1-x^4))-1],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Sep 08 2019 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4))-1)))

Formula

a(n+16) = (n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8)*(n+9)*(n+10)*(n+11)*(n+12)*(n+13)*(n+14)*(n+15)*n*a(n) - (n+15)*(n+14)*(n+13)*(n+12)*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+6)*(n+5)*(n+4)*(n+3)*(n+2)*a(n+2) - 2*(n+15)*(n+14)*(n+13)*(n+12)*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+6)*(n+5)*(n+4)*(n+3)*a(n+3) - 2*(n+15)*(n+14)*(n+13)*(n+12)*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+6)*(n+5)*(n+4)*a(n+4) + 2*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+6)*(n+5)*(n+15)*(n+14)*(n+13)*(n+12)*a(n+5) + 3*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+6)*(n+15)*(n+14)*(n+13)*(n+12)*a(n+6) + 4*(n+11)*(n+10)*(n+9)*(n+8)*(n+7)*(n+15)*(n+14)*(n+13)*(n+12)*a(n+7) - 4*(n+11)*(n+10)*(n+9)*(n+15)*(n+14)*(n+13)*(n+12)*a(n+9) - (n+15)*(n+14)*(n+13)*(n+12)*(n+11)*(3*n+20)*a(n+10) - (2*n+11)*(n+15)*(n+14)*(n+13)*(n+12)*a(n+11) + 2*(n+19)*(n+15)*(n+14)*(n+13)*a(n+12) + 2*(n+15)*(n+14)*(n+17)*a(n+13) + (n+18)*(n+15)*a(n+14) + a(n+15). - Robert Israel, Mar 12 2020
a(n) ~ exp(-68413/92160 + 295*n^(1/5) / (192*6^(4/5)) + 15*3^(2/5)*n^(2/5) / (32*2^(3/5)) + 5*n^(3/5) / (4*6^(2/5)) + 5*n^(4/5) / (4*6^(1/5)) - n) * n^(n - 1/10) / (sqrt(5) * 6^(1/10)) * (1 + 18025/(18432*6^(1/5)*n^(1/5))). - Vaclav Kotesovec, Dec 02 2021

A294216 E.g.f.: exp(1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)) - 1).

Original entry on oeis.org

1, 1, 5, 31, 265, 2621, 30901, 411475, 6232241, 103992985, 1909517221, 38038551431, 819452533945, 18924054045781, 466804248205205, 12234394315501051, 339537099805667041, 9941204417393228465, 306212116909615474501, 9894769026271957204975
Offset: 0

Views

Author

Seiichi Manyama, Oct 25 2017

Keywords

Crossrefs

Column k=5 of A294212.

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Exp[1/Times@@(1-x^Range[5])-1],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Sep 08 2018 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5))-1)))

A294213 E.g.f.: exp(1/((1-x)*(1-x^2)) - 1).

Original entry on oeis.org

1, 1, 5, 25, 193, 1601, 16741, 190345, 2509025, 35825473, 569012581, 9716400761, 180303804385, 3569527588225, 75681964322693, 1700163418683241, 40499757023856961, 1016190431274596225, 26843084299482509125, 743180975111364212953
Offset: 0

Views

Author

Seiichi Manyama, Oct 25 2017

Keywords

Crossrefs

Column k=2 of A294212.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[E^(1/((1-x)*(1-x^2)) - 1), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 26 2017 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2))-1)))

Formula

a(n) ~ exp(-61/96 + 3*n^(1/3)/4 + 3*n^(2/3)/2 - n) * n^(n - 1/6) / sqrt(3) * (1 + 25/(64*n^(1/3))). - Vaclav Kotesovec, Oct 26 2017
a(n) = n*a(n-1) + (n-1)*(2*n - 1)*a(n-2) - 2*(n-3)*(n-2)*(n-1)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4) + (n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5). - Vaclav Kotesovec, Dec 02 2021
From Peter Bala, Oct 17 2023: (Start)
a(n+k) == a(n) (mod k) for all n and k >= 1. Hence for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. Cf. A047974.
a(5*n + 2) == a(5*n + 3) == 0 (mod 5);
a(25*n + 3) == a(25*n + 8) == a(25*n + 13) == a(25*n + 17) == a(25*n + 18) == a(25*n + 23) == 0 (mod 5^2);
a(125*n + 18) == a(125*n + 67) == a(125*n + 93) == a(125*n + 118) == 0 (mod 5^3). (End)
Showing 1-6 of 6 results.