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.

A101028 Numerator of partial sums of a certain series. First member (m = 2) of a family.

Original entry on oeis.org

1, 11, 79, 479, 5297, 69071, 69203, 471181, 8960447, 44831407, 1031626241, 5160071143, 15484789693, 64166447971, 1989542332021, 3979714828967, 27861681000449, 1030996803010973, 1031094241305773, 42278288849598913, 1818093633186492859, 1818204269645957299, 85460151199040573933
Offset: 1

Views

Author

Wolfdieter Lang, Dec 17 2004

Keywords

Comments

The denominators are given in A101029.
The limit s = lim_{n -> infinity} s(n) with s(n) defined below equals 3*Sum_{k >= 1} zeta(2*k+1)/2^(2*k) with Euler's (or Riemann's) zeta function. This limit is 3*(2*log(2)-1) = 1.158883083...; see the Abramowitz-Stegun reference p. 259, eq. 6.3.15 with z = 1/2 together with p. 258, eqs. 6.3.5 and 6.3.3.
This is the first member (m = 2) of a family of rational partial sum sequences s(n,m) = (m-1)*m*(m+1)*Sum_{k = 1..n} 1/((m*k-1)*(m*k)*(m*k+1)) which have limit s(m) = lim_{n -> infinity} s(n,m) = -(gamma + Psi(1/m) + m/2 + Pi*cot(Pi*x)/2), with the Euler-Mascheroni constant gamma and the digamma function Psi. The same limit is reached by (m^2-1)*Sum_{k >= 0} zeta(2*k+1)/m^(2*k).
From Peter Bala, Feb 17 2022: (Start)
Let F(n) = (6*n/(2*n-1))*( 1/(1*2)*(n-1)/n - 1/(2*3)*(n-1)*(n-2)/(n*(n+1)) + 1/(3*4)*(n-1)*(n-2)*(n-3)/(n*(n+1)*(n+2)) - 1/(4*5)*(n-1)*(n-2)*(n-3)*(n-4)/(n*(n+1)*(n+2)*(n+3)) + ...). Then F(n+1) = 6*Sum_{k = 1..n} 1/((2*k-1)*(2*k)*(2*k+1)). Cf. A082687.
This identity allows us to extend the definition of Sum_{k = 1..n} 1/((2*k-1)*(2*k)*(2*k+1)) to non-integral values of n. (End)

Examples

			s(3)= 6*(1/(1*2*3)+ 1/(3*4*5) + 1/(5*6*7)) = 79/70, hence a(3)=79 and A101029(3)=70.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 258-259.
  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Crossrefs

Cf. A101627 (m=3), A101629 (m=4), A101631 (m=5).
Cf. A082687, A101029 (denominators).

Programs

  • PARI
    a(n) = numerator(6*sum(k=1, n, 1/((2*k-1)*(2*k)*(2*k+1)))); \\ Michel Marcus, Feb 28 2022

Formula

a(n) = numerator(s(n)) with s(n) = 6*Sum_{k = 1..n} 1/((2*k-1)*(2*k)*(2*k+1)).

Extensions

More terms from Michel Marcus, Feb 28 2022

A375921 a(n) = LCM(1,2, ..., 2n+1)/6.

Original entry on oeis.org

1, 10, 70, 420, 4620, 60060, 60060, 2042040, 38798760, 38798760, 892371480, 4461857400, 13385572200, 388181593800, 12033629407800, 24067258815600, 24067258815600, 890488576177200, 890488576177200, 36510031623265200, 1569931359800403600, 1569931359800403600
Offset: 1

Views

Author

Gregory Gerard Wojnar, Sep 02 2024

Keywords

Comments

Agrees largely with A101029. The first difference is at n=8 for which a(8)/A101029(8) = 5. There are also other differences; the ratio of the two series entries appears to be always an integer.

Crossrefs

Programs

  • Mathematica
    a[n_]:= LCM@@Range[1,2n+1]/6; Array[a,22] (* Stefano Spezia, Sep 02 2024 *)
  • PARI
    a(n) = lcm([1..2*n+1])/6; \\ Michel Marcus, Sep 02 2024
    
  • Python
    from math import lcm
    def A375921(n): return lcm(*range(1,n+1<<1))//6 # Chai Wah Wu, Sep 26 2024

Formula

a(n) = A003418(2n+1)/6.
Showing 1-2 of 2 results.