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 1-4 of 4 results.

A320080 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. 1/(1 - k*log(1 + x)).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 2, 0, 1, 4, 15, 28, 4, 0, 1, 5, 28, 114, 172, 14, 0, 1, 6, 45, 296, 1152, 1328, 38, 0, 1, 7, 66, 610, 4168, 14562, 12272, 216, 0, 1, 8, 91, 1092, 11020, 73376, 220842, 132480, 600, 0, 1, 9, 120, 1778, 24084, 248870, 1550048, 3907656, 1633344, 6240, 0
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 05 2018

Keywords

Examples

			E.g.f. of column k: A_k(x) = 1 + k*x/1! + k*(2*k - 1)*x^2/2! + 2*k*(3*k^2 - 3*k + 1)*x^3/3! + 2*k*(12*k^3 - 18*k^2 + 11*k - 3)*x^4/4! + ...
Square array begins:
  1,   1,     1,      1,      1,       1,  ...
  0,   1,     2,      3,      4,       5,  ...
  0,   1,     6,     15,     28,      45,  ...
  0,   2,    28,    114,    296,     610,  ...
  0,   4,   172,   1152,   4168,   11020,  ...
  0,  14,  1328,  14562,  73376,  248870,  ...
		

Crossrefs

Columns k=0..5 give A000007, A006252, A088501, A335531, A354147, A365604.
Main diagonal gives A317172.

Programs

  • Mathematica
    Table[Function[k, n! SeriesCoefficient[1/(1 - k Log[1 + x]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

Formula

E.g.f. of column k: 1/(1 - k*log(1 + x)).
A(n,k) = Sum_{j=0..n} Stirling1(n,j)*j!*k^j.
A(0,k) = 1; A(n,k) = k * Sum_{j=1..n} (-1)^(j-1) * (j-1)! * binomial(n,j) * A(n-j,k). - Seiichi Manyama, May 22 2022

A354264 Expansion of e.g.f. 1/(1 + 4 * log(1-x)).

Original entry on oeis.org

1, 4, 36, 488, 8824, 199456, 5410208, 171209664, 6192052800, 251937937920, 11389639660032, 566394573855744, 30726758349800448, 1805828538127687680, 114293350061315678208, 7750480651439579529216, 560615413313367534698496, 43085423893717998388740096
Offset: 0

Views

Author

Seiichi Manyama, May 21 2022

Keywords

Crossrefs

Column k=4 of A320079.

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/(1+4*log(1-x))))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=4*sum(j=1, i, (j-1)!*binomial(i, j)*v[i-j+1])); v;
    
  • PARI
    a(n) = sum(k=0, n, 4^k*k!*abs(stirling(n, k, 1)));

Formula

E.g.f.: 1/(1 + 4 * log(1-x)).
a(0) = 1; a(n) = 4 * Sum_{k=1..n} (k-1)! * binomial(n,k) * a(n-k).
a(n) = Sum_{k=0..n} 4^k * k! * |Stirling1(n, k)|.
a(n) ~ n! * exp(n/4) / (4 * (exp(1/4) - 1)^(n+1)). - Vaclav Kotesovec, Jun 04 2022

A354751 Expansion of e.g.f. 1 / (1 - log(1 + 4*x) / 4).

Original entry on oeis.org

1, 1, -2, 14, -152, 2264, -42832, 982512, -26484096, 820207488, -28692711168, 1118821622016, -48112717347840, 2261868010650624, -115400220781209600, 6350152838136428544, -374874781697133871104, 23632196147497381625856, -1584445791263626895228928
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 06 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - Log[1 + 4 x]/4), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS1[n, k] k! 4^(n - k), {k, 0, n}], {n, 0, 18}]
  • PARI
    my(x='x + O('x^20)); Vec(serlaplace(1/(1-log(1+4*x)/4))) \\ Michel Marcus, Jun 06 2022

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * k! * 4^(n-k).
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * (k-1)! * (-4)^(k-1) * a(n-k).

A365600 Expansion of e.g.f. 1 / (1 - 4 * log(1 + x))^(3/4).

Original entry on oeis.org

1, 3, 18, 174, 2292, 38292, 774624, 18399840, 501868416, 15456483840, 530462128896, 20073406663296, 830293158570624, 37267057695192192, 1803930663341528064, 93672204405378891264, 5193925606670524254720, 306280622206497897745920
Offset: 0

Views

Author

Seiichi Manyama, Sep 11 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Product[4*j + 3, {j, 0, k - 1}] * StirlingS1[n, k], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Sep 13 2023 *)
  • PARI
    a(n) = sum(k=0, n, prod(j=0, k-1, 4*j+3)*stirling(n, k, 1));

Formula

a(n) = Sum_{k=0..n} (Product_{j=0..k-1} (4*j+3)) * Stirling1(n,k).
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (4 - k/n) * (k-1)! * binomial(n,k) * a(n-k).
a(n) ~ Gamma(1/4) * n^(n + 1/4) / (2^(3/2) * sqrt(Pi) * (exp(1/4) - 1)^(n + 3/4) * exp(n - 3/16)). - Vaclav Kotesovec, Nov 10 2023
Showing 1-4 of 4 results.