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.

A197070 Decimal expansion of the Dirichlet eta-function at 3.

Original entry on oeis.org

9, 0, 1, 5, 4, 2, 6, 7, 7, 3, 6, 9, 6, 9, 5, 7, 1, 4, 0, 4, 9, 8, 0, 3, 6, 2, 1, 1, 3, 3, 5, 8, 7, 4, 9, 3, 0, 7, 3, 7, 3, 9, 7, 1, 9, 2, 5, 5, 3, 7, 4, 1, 6, 1, 3, 4, 4, 2, 0, 3, 6, 6, 6, 5, 0, 6, 3, 7, 8, 6, 5, 4, 3, 3, 9
Offset: 0

Views

Author

R. J. Mathar, Oct 09 2011

Keywords

Comments

This constant is irrational by Apéry's theorem. - Charles R Greathouse IV, Feb 11 2024

Examples

			0.9015426773696957140498036211335874930737...
		

Crossrefs

Cf. A002117 (zeta(3)), A058312, A058313, A072691, A136675, A233090 (5*zeta(3)/8), A233091 (7*zeta(3)/8), A334582.

Programs

Formula

Equals 3*zeta(3)/4 = 3*A002117/4.
Also equals the integral over the unit cube [0,1]x[0,1]x[0,1] of 1/(1+x*y*z) dx dy dz. - Jean-François Alcover, Nov 24 2014
Equals Sum_{n>=1} (-1)^(n+1)/n^3. - Terry D. Grant, Aug 03 2016
Equals Lim_{n -> infinity} A136675(n)/A334582(n). - Petros Hadjicostas, May 07 2020
Equals Sum_{n>=1} AH(2*n)/n^2, where AH(n) = Sum_{k=1..n} (-1)^(k+1)/k = A058313(n)/A058312(n) is the n-th alternating harmonic number (Stewart, 2020). - Amiram Eldar, Oct 04 2021
Equals -int_0^1 log(x)log(1+x)/x dx [Barbieri] - R. J. Mathar, Jun 07 2024

A136675 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^3.

Original entry on oeis.org

1, 7, 197, 1549, 195353, 194353, 66879079, 533875007, 14436577189, 14420574181, 19209787242911, 19197460851911, 42198121495296467, 6025866788581781, 6027847576222613, 48209723660000029, 236907853607882606477
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

a(n) is prime for n in A136683.
Lim_{n -> infinity} a(n)/A334582(n) = A197070. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 7/8, 197/216, 1549/1728, 195353/216000, 194353/216000, 66879079/74088000, 533875007/592704000, ... = a(n)/A334582(n). - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Programs

  • Maple
    map(numer,ListTools:-PartialSums([seq((-1)^(k+1)/k^3, k=1..100)])); # Robert Israel, Nov 09 2023
  • Mathematica
    (* Program #1 *) Table[Numerator[Sum[(-1)^(k+1)/k^3, {k,1,n}]], {n,1,50}]
    (* Program #2 *) Numerator[Accumulate[Table[(-1)^(k+1) 1/k^3, {k,50}]]] (* Harvey P. Dale, Feb 12 2013 *)
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^3)); \\ Michel Marcus, May 07 2020

A195506 Denominator of Sum_{k=1..n} H(k)/k^2, where H(k) is the k-th harmonic number.

Original entry on oeis.org

1, 8, 216, 1728, 216000, 216000, 74088000, 592704000, 16003008000, 16003008000, 21300003648000, 21300003648000, 46796108014656000, 46796108014656000, 46796108014656000, 374368864117248000, 1839274229408039424000, 1839274229408039424000
Offset: 1

Views

Author

Franz Vrabec, Sep 19 2011

Keywords

Comments

Lim_{n -> infinity} (A195505(n)/a(n)) = 2*Zeta(3) [L. Euler].
For n = 1 to n = 13, a(n) = A334582(n), but a(14) = 46796108014656000 <> 6685158287808000 = A334582(14). - Petros Hadjicostas, May 06 2020

Examples

			a(2) = 8 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. A002117, A195505 (numerators), A334582.

Programs

  • Mathematica
    s = 0; Table[s = s + HarmonicNumber[n]/n^2; Denominator[s], {n, 20}] (* T. D. Noe, Sep 20 2011 *)
  • PARI
    H(n) = sum(k=1, n, 1/k);
    a(n) = denominator(sum(k=1, n, H(k)/k^2)); \\ Michel Marcus, May 07 2020
Showing 1-3 of 3 results.