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.

A345682 a(n) = n! * Sum_{k=1..n} 1/(k*floor(n/k)).

Original entry on oeis.org

1, 2, 7, 26, 148, 804, 6228, 47424, 441936, 4288320, 50437440, 560373120, 7723935360, 106618256640, 1614841401600, 25127582054400, 446784010444800, 7727747269939200, 152873884406476800, 2966599550251008000, 62987912790921216000, 1378192085174919168000
Offset: 1

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n! * Sum[1/(k*Floor[n/k]), {k, 1, n}], {n, 1, 25}]
    Table[n! * Sum[(HarmonicNumber[Floor[n/j]] - HarmonicNumber[Floor[n/(1 + j)]])/j, {j, 1, n}], {n, 1, 25}]
  • PARI
    a(n) = n!*sum(k=1, n, 1/(k*(n\k))); \\ Michel Marcus, Jun 24 2021
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k)/k)/(1-x))) \\ Seiichi Manyama, Jul 23 2022

Formula

a(n) ~ c * n!, where c = Sum_{j>=1} log(1 + 1/j)/j = A131688 = 1.25774...
E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k)/k. - Seiichi Manyama, Jul 23 2022

A345684 a(n) = n! * Sum_{k=1..n} k/floor(n/k).

Original entry on oeis.org

1, 5, 32, 198, 1584, 12480, 122520, 1214640, 14011200, 166924800, 2274894720, 31135104000, 485667705600, 7710089587200, 133974352512000, 2386854434764800, 46621903994265600, 918384939343872000, 19760215067873280000, 430137075045629952000, 10042411264251125760000
Offset: 1

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n!*Sum[k/Floor[n/k], {k, 1, n}], {n, 1, 25}]
    Table[n!*Sum[(Floor[n/j] - Floor[n/(1 + j)])*((1 + Floor[n/j] + Floor[n/(1 + j)])/2/j), {j, 1, n}], {n, 1, 25}]
  • PARI
    a(n) = n!*sum(k=1, n, k/(n\k)); \\ Michel Marcus, Jun 23 2021
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, k*(1-x^k)*log(1-x^k))/(1-x))) \\ Seiichi Manyama, Jul 23 2022

Formula

a(n) ~ c * n^2 * n!, where c = Sum_{j>=1} (2*j + 1) / (2*j^3*(j+1)^2) = Pi^2/12 + zeta(3)/2 - 1 = 0.423495...
E.g.f.: -(1/(1-x)) * Sum_{k>0} k * (1 - x^k) * log(1 - x^k). - Seiichi Manyama, Jul 23 2022

A355987 a(n) = n! * Sum_{k=1..n} 1/floor(n/k)!.

Original entry on oeis.org

1, 3, 13, 61, 421, 2641, 23521, 203281, 2071441, 22407841, 286403041, 3453468481, 51122111041, 759194916481, 12216117513601, 203300293996801, 3811792426041601, 69634723878720001, 1444704854104512001, 29725332567567436801, 658231789483184716801
Offset: 1

Views

Author

Seiichi Manyama, Jul 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := n! * Sum[1/Floor[n/k]!, {k, 1, n}]; Array[a, 21] (* Amiram Eldar, Jul 22 2022 *)
  • PARI
    a(n) = n!*sum(k=1, n, 1/(n\k)!);
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(sum(k=1,N, (1-x^k)*(exp(x^k)-1))/(1-x)))

Formula

E.g.f.: (1/(1-x)) * Sum_{k>0} (1 - x^k) * (exp(x^k) - 1).
a(n) ~ c * n! * n, where c = 0.59962032... - Vaclav Kotesovec, Aug 03 2022
Conjecture: c = Sum_{k>=1} 1/((k+1)!*k) = 2 - exp(1) - A001620 + A091725. - Vaclav Kotesovec, Sep 24 2023

A356011 a(n) = n! * Sum_{k=1..n} 1/(k! * floor(n/k)).

Original entry on oeis.org

1, 2, 6, 17, 80, 337, 2240, 14681, 117010, 1023941, 10900472, 108881665, 1375544846, 17732140805, 247041590476, 3605768497217, 59990390084690, 977383707751621, 18214603019184800, 337615168055209601, 6763842079452393622, 141262515443311046885
Offset: 1

Views

Author

Seiichi Manyama, Jul 23 2022

Keywords

Crossrefs

Row sums of A356013.

Programs

  • Mathematica
    Table[n! * Sum[1/(k!*Floor[n/k]), {k,1,n}], {n,1,25}] (* Vaclav Kotesovec, Aug 11 2025 *)
  • PARI
    a(n) = n!*sum(k=1, n, 1/(k!*(n\k)));
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k)/k!)/(1-x)))

Formula

E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k)/k!.
a(n) ~ exp(1) * (n-1)!. - Vaclav Kotesovec, Aug 11 2025

A345686 a(n) = n! * Sum_{k=1..n} n/floor(n/k)^2.

Original entry on oeis.org

1, 5, 38, 222, 1974, 14640, 154580, 1476720, 17753400, 205430400, 2924030592, 38559628800, 623916216000, 9701871379200, 172359487872000, 3007238402488320, 60362232844193280, 1161408374590464000, 25603215951785472000, 547592177551491072000, 12990145748633044992000
Offset: 1

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Cf. A345683.

Programs

  • Mathematica
    Table[n! * Sum[n/Floor[n/k]^2, {k, 1, n}], {n, 1, 25}]
    Table[n*n!*(Sum[(Floor[n/j] - Floor[n/(j + 1)])/j^2, {j, 1, n}]), {n, 1, 25}]

Formula

a(n) ~ c * n^2 * n!, where c = Sum_{j>-1} 1/(j^3*(j+1)) = zeta(3) - Pi^2/6 + 1.

A342933 a(n) = n! * Sum_{k=1..n} k^2/floor(n/k).

Original entry on oeis.org

1, 9, 80, 654, 6264, 59760, 665160, 7565040, 96929280, 1286046720, 18976083840, 286363123200, 4775047200000, 81792956044800, 1515077749785600, 28898470215014400, 594066352700620800, 12467555729620992000, 280797084422959104000, 6460327992512249856000, 157769680941941612544000
Offset: 1

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Comments

In general, for m>=0, Sum_{k=1..n} k^m / floor(n/k) ~ n^(m+1) * (-1 + Sum_{j=2..m+2} zeta(j) / (m+1)).

Crossrefs

Programs

  • Mathematica
    Table[n!*Sum[k^2/Floor[n/k], {k, 1, n}], {n, 1, 25}]
    Table[n!*Sum[(Floor[n/j]*(1 + Floor[n/j])*(1 + 2*Floor[n/j]) - Floor[n/(1 + j)]*(1 + Floor[n/(1 + j)])*(1 + 2*Floor[n/(1 + j)]))/6/j, {j, 1, n}], {n, 1, 25}]

Formula

a(n) ~ c * n^3 * n!, where c = Sum_{j>=1} (1 + 3*j*(j+1)) / (3*j^4*(j+1)^3) = (zeta(4) + zeta(3) + zeta(2))/3 - 1 = Pi^2/18 + Pi^4/270 + zeta(3)/3 - 1.
Showing 1-6 of 6 results.