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 1-2 of 2 results.

A334582 Denominator of Sum_{k=1..n} (-1)^(k+1)/k^3.

Original entry on oeis.org

1, 8, 216, 1728, 216000, 216000, 74088000, 592704000, 16003008000, 16003008000, 21300003648000, 21300003648000, 46796108014656000, 6685158287808000, 6685158287808000, 53481266302464000, 262753461344005632000, 262753461344005632000
Offset: 1

Views

Author

Petros Hadjicostas, May 06 2020

Keywords

Comments

For n = 1 to n = 13, a(n) = A195506(n), but a(14) = 6685158287808000 <> 46796108014656000 = A195506(14).
Lim_{n -> infinity} A136675(n)/a(n) = A197070.

Examples

			The first few fractions are 1, 7/8, 197/216, 1549/1728, 195353/216000, 194353/216000, 66879079/74088000, 533875007/592704000, ... = A136675/A334582.
		

Crossrefs

Cf. A136675 (numerators), A195506, A197070.

Programs

  • Maple
    b := proc(n) local k: add((-1)^(k + 1)/k^3, k = 1 .. n): end proc:
    seq(denom(b(n)), n=1..30);
  • Mathematica
    Denominator @ Accumulate[Table[(-1)^(k + 1)/k^3, {k, 1, 18}]] (* Amiram Eldar, May 08 2020 *)
  • PARI
    a(n) = denominator(sum(k=1, n, (-1)^(k+1)/k^3)); \\ Michel Marcus, May 07 2020

Extensions

Offset changed to 1 by Georg Fischer, Jul 13 2023

A195505 Numerator of Sum_{k=1..n} H(k)/k^2, where H(k) is the k-th harmonic number.

Original entry on oeis.org

1, 11, 341, 2953, 388853, 403553, 142339079, 1163882707, 31983746689, 32452469713, 43725835522403, 44184852180503, 97954699428176291, 98731028315167091, 99421162547987123, 800313205356878959, 3953829021224881128767, 3973669953994085875967
Offset: 1

Views

Author

Franz Vrabec, Sep 19 2011

Keywords

Comments

Lim_{n-> infinity} (a(n)/A195506(n)) = 2*Zeta(3) [L. Euler].
Sum_{k = 1..n} H(k)/k^2 is an example of a multiple harmonic (star) sum. Euler's result Sum_{k = 1..inf} H(k)/k^2 = 2*zeta(3) was the first evaluation of a multiple zeta star value. - Peter Bala, Jan 31 2019

Examples

			a(2) = 11 because 1 + (1 + 1/2)/2^2 = 11/8.
The first few fractions are 1, 11/8, 341/216, 2953/1728, 388853/216000, 403553/216000, 142339079/74088000, 1163882707/592704000, ... = A195505/A195506. - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Cf. A001008, A002117, A036970, A195506 (denominators).

Programs

  • Mathematica
    s = 0; Table[s = s + HarmonicNumber[n]/n^2; Numerator[s], {n, 20}] (* T. D. Noe, Sep 20 2011 *)
  • PARI
    H(n) = sum(k=1, n, 1/k);
    a(n) = numerator(sum(k=1, n, H(k)/k^2)); \\ Michel Marcus, May 07 2020

Formula

From Peter Bala, Jan 31 2019: (Start)
Let S(n) = Sum_{k = 1..n} H(k)/k^2. Then
S(n) = 1 + (1 + 1/2^3)*(n-1)/(n+1) + (1/2^3 + 1/3^3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3^3 + 1/4^3)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...
S(n)/n = 1 + (1/2^4 - 1)*(n-1)/(n+1) + (1/3^4 - 1/2^4)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^4 - 1/3^4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...
For odd n >= 3, 1/2*S((n-1)/2) = (n-1)/(n+1) + 1/2^3*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3^3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ....
Cf. A001008. See the Bala link in A036970. (End)
Showing 1-2 of 2 results.