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 1-3 of 3 results.

A355203 E.g.f. A(x) satisfies A'(x) = 1 + A(1 - exp(-x)).

Original entry on oeis.org

1, 1, 0, -2, 4, 10, -150, 838, 222, -82616, 1408364, -13862308, -18747672, 5307622274, -170657860276, 3561218897884, -33756455501714, -1481233045213718, 116803294574962288, -5108843717328225572, 157037998518149186728, -1976107915155933805542
Offset: 1

Views

Author

Seiichi Manyama, Jun 24 2022

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n)); v[1]=1; for(i=1, n-1, v[i+1]=sum(j=1, i, (-1)^(i-j)*stirling(i, j, 2)*v[j])); v;

Formula

a(1) = 1; a(n+1) = Sum_{k=1..n} (-1)^(n-k) * Stirling2(n,k) * a(k).

A355204 E.g.f. A(x) satisfies A'(x) = 1 + 2 * A(log(1+x)).

Original entry on oeis.org

1, 2, 2, -4, 0, 68, -588, 2728, 17688, -766960, 14239512, -164672640, -494840680, 109461302008, -4446935274696, 122761839873664, -1889647100968176, -50347073461051088, 6582857386505201520, -397095020380174033424, 17279075935957171412288
Offset: 1

Views

Author

Seiichi Manyama, Jun 24 2022

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n)); v[1]=1; for(i=1, n-1, v[i+1]=2*sum(j=1, i, stirling(i, j, 1)*v[j])); v;

Formula

a(1) = 1; a(n+1) = 2 * Sum_{k=1..n} Stirling1(n,k) * a(k).

A355208 E.g.f. A(x) satisfies A'(x) = 1 + A(2 * log(1+x)).

Original entry on oeis.org

1, 2, 6, 28, 236, 4400, 197552, 20430656, 4600591488, 2179887358272, 2130534442932416, 4243581375963409024, 17097951082212352465536, 138722374358947243721661440, 2260145794657531151029628653568, 73822509077371344216463442074629120
Offset: 1

Views

Author

Seiichi Manyama, Jun 24 2022

Keywords

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n)); v[1]=1; for(i=1, n-1, v[i+1]=sum(j=1, i, 2^j*stirling(i, j, 1)*v[j])); v;

Formula

a(1) = 1; a(n+1) = Sum_{k=1..n} 2^k * Stirling1(n,k) * a(k).
Showing 1-3 of 3 results.