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.

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A304589 a(n) = (n!)^3 * Sum_{k=1..n-1} 1/(k*(n-k))^3.

Original entry on oeis.org

0, 0, 8, 54, 1240, 70000, 7941968, 1589632128, 512918521344, 249820864339968, 174720109813751808, 168721560082538496000, 217977447876560510976000, 367117517435096337481728000, 788739873984137255456342016000, 2122296978948474538763602624512000
Offset: 0

Views

Author

Vaclav Kotesovec, May 15 2018

Keywords

Comments

In general, for m > 1, Sum_{k=1..n-1} 1/(k*(n-k))^m is asymptotic to 2*Zeta(m)/n^m.

Crossrefs

Programs

  • GAP
    List([0..20],n->Factorial(n)^3*Sum([1..n-1],k->1/(k*(n-k))^3)); # Muniru A Asiru, May 16 2018
  • Maple
    seq(factorial(n)^3*add(1/(k*(n-k))^3,k=1..n-1),n=0..20); # Muniru A Asiru, May 16 2018
  • Mathematica
    Table[n!^3*Sum[1/(k*(n-k))^3, {k, 1, n-1}], {n, 0, 20}]
    CoefficientList[Series[PolyLog[3, x]^2, {x, 0, 20}], x] * Range[0,20]!^3

Formula

Recurrence: n^2*(12*n^4 - 108*n^3 + 354*n^2 - 501*n + 260)*a(n) = 2*(n-1)^2*(24*n^7 - 306*n^6 + 1620*n^5 - 4599*n^4 + 7516*n^3 - 7015*n^2 + 3444*n - 696)*a(n-1) - 6*(n-2)^5*(12*n^7 - 162*n^6 + 906*n^5 - 2700*n^4 + 4583*n^3 - 4378*n^2 + 2163*n - 436)*a(n-2) + 2*(n-3)^5*(n-2)^3*(24*n^7 - 342*n^6 + 2004*n^5 - 6201*n^4 + 10816*n^3 - 10497*n^2 + 5208*n - 1048)*a(n-3) - (n-4)^6*(n-3)^5*(n-2)^3*(12*n^4 - 60*n^3 + 102*n^2 - 69*n + 17)*a(n-4).
a(n) / (n!)^3 ~ 2*Zeta(3)/n^3.

A304654 a(n) = (n!)^2 * Sum_{k=1..n-1} 1/(k^2*(n-k)).

Original entry on oeis.org

0, 0, 4, 27, 328, 6500, 192216, 7952112, 438941952, 31185057024, 2772643115520, 301622403456000, 39413353102848000, 6091955683706880000, 1099401414283210752000, 229088914497045356544000, 54589580461769879715840000, 14750581694440372638842880000
Offset: 0

Views

Author

Vaclav Kotesovec, May 16 2018

Keywords

Crossrefs

Programs

  • Mathematica
    Table[(n!)^2 * Sum[1/(k^2*(n-k)), {k, 1, n-1}], {n, 0, 20}]

Formula

Recurrence: (2*n - 3)*a(n) = (6*n^3 - 25*n^2 + 33*n - 12)*a(n-1) - (n-2)^2*(6*n^3 - 29*n^2 + 42*n - 15)*a(n-2) + (n-3)^3*(n-2)^3*(2*n - 1)*a(n-3).
a(n)/(n!)^2 ~ Pi^2/(6*n).
Showing 1-2 of 2 results.