cp's OEIS Frontend

This sequence coincides with the sequence f(n) = denominator of 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) iff n <= 38. But a(39) = 6414924694381721303722858446525, f(39) = 583174972216520118520259858775. - T. D. Noe, Aug 04 2004 [See A350670(n-1).]

Coincides for n=1..42 with the denominators of a series for Pi*sqrt(2)/4 and then starts to differ. See A127676.

a(floor((n+1)/2)) = gcd(a(n), A051426(n)). - Reinhard Zumkeller, Apr 25 2011

A051417(n) = a(n+1)/a(n).

For integers a and b, H(n,a,b) is the sum of the fractions 1/(a i + b), i = 0,1,..,n-1. This database already contains six instances of generalized harmonic numbers. Partial sums of the harmonic series 1+1/2+1/3+1/4+... are given by the sequence of harmonic numbers H(n,1,1) = A001008(n) / A002805(n).

The Jeep problem gives rise to the series H(n,2,1) = A025550(n) / A025547(n). Recent additions to the database are 3 * H(n,3,1) = A074596(n) / A051536(n), 3 * H(n,3,2) = A074597(n) / A051540(n), 4 * H(n,4,1) = A074598(n) / A051539(n) and 4 * H(n,4,3) = A074637(n) / A074638(n) . The numerator of H(n,4,1) is A075136. The fractions H(n,5,1), H(n,5,2), H(n,5,3) and H(n,5,4) are in A075137-A075144.

The sequence H(n,a,b) is of interest only when a and b are relatively prime. The sequence can also be computed as H(n,a,b) = (PolyGamma[n+1+b/a] - PolyGamma[1+b/a])/a. The sequence H(n,a,b) diverges for all a and b.

According to Hardy and Wright, if p is an odd prime, then p divides the numerator of the harmonic number H(p-1,1,1). This result can be extended to generalized harmonic numbers: for odd integer n, let q = (n-2)a + 2b. If q is prime, then q divides the numerator of H(n-1,a,b). For this sequence (a=3, b=1) we conclude that 11 divides a(4), 17 divides a(6), 29 divides a(10) and 47 divides a(16).

Graham, Knuth and Patashnik define another type of generalized harmonic number as the sum of fractions 1/i^k, i=1,...,n. For k=2, the sequence of fractions is A007406(n) / A007407(n).

a((p-1)/2) is divisible by prime p > 3.

The limit of the rationals r(n) = Sum_{k=1..n} 1/(2*k-1)^2, for n -> infinity, is (Pi^2)/8 = (1 - 1/2^2)*Zeta(2), which is approximately 1.233700550.

r(n) = (Psi(1, 1/2) - Psi(1, n+1/2))/4 for n >= 1, where Psi(n,k) = Polygamma(n,k) is the n-th derivative of the digamma function. Psi(1, 1/2) = 3*Zeta(2) = Pi^2/2. - Jean-François Alcover, Dec 02 2013 [Corrected by Petros Hadjicostas, May 09 2020]

The recurrence given by Cserti (2000), page 5, (32) is used to calculate the resistance between two arbitrarily spaced nodes in an infinite square lattice whose edges are replaced by one-ohm resistors. The lower triangle, including the diagonal, in Table I of Atkinson and Steenwijk (1999), page 487, is reproduced. The solution to the resistor grid problem shown in the xkcd Web Comic #356 "Nerd Sniping", provided in A211074, is the special case (j,k) = (2,1).

Using the terms of A280079 and A280317 as pairs of grid indices leads to strictly increasing resistances, i.e., R(A280079(m),A280317(m)) > R(A280079(i),A280317(i)) for m > i. This implies that for grid points on the same radius the resistance increases with the circumferential angle between 0 and Pi/4. The further dependence of the resistance along the circumferential angle with a fixed radius results from symmetry. - Hugo Pfoertner, Aug 31 2022

2*(1 + 1/3 + ... + 1/(2*n-1))/Pi = a(n)/(A350670(n)*Pi) is the equivalent resistance between the points (0,0) and (n,n) on a 2-dimension infinite square grid of unit resistors. - Jianing Song, Apr 28 2025

|a(n)| = numerator of Sum_{k=1..n} 1/(n*(2*k-1)).

Let f(x) = (1/2)*log((1+sqrt(x))/(1-sqrt(x))) and c(n) = Integral_{x=0..1} f(x)*x^(n-1) dx, then for n>=1, c(n) = |a(n+1)|/A071968(n) and (f(x))^2 = Sum_{n>=1} c(n)*x^n. - Groux Roland, Dec 14 2010

For the denominators see A350670.

This sequence coincides with A025550(n+1), for n = 0, 1, ..., 37. See the comments there.

Thanks to Ralf Steiner for sending me a paper where this and similar sums appear.

For the numerators see A350669.

This sequence coincides with A025547(n+1), for n = 0, 1, ..., 37. See the comments there.

Thanks to Ralf Steiner for sending me a paper where this and similar sums appear.

a((p-1)/2) is divisible by prime p > 5.

Denominators are in A128493.

The limit of the rationals r(n) = Sum_{k=1..n} 1/(2k-1)^4, for n -> infinity, is (Pi^4)/96 = (1 - 1/2^4)*zeta(4), which is approximately 1.014678032.

r(n) = (Psi(3, 1/2) - Psi(3, n+1/2))/(3!*2^4) for n >= 1, where Psi(n,k) = Polygamma(n,k) is the n^th derivative of the digamma function. Psi(3, 1/2) = 3!*15*zeta(4) = Pi^4. - Jean-François Alcover, Dec 02 2013

Warning: Usually, Theta3(x) = Sum_{n=-oo..+oo} x^(n^2). - Joerg Arndt, Mar 31 2024

The denominators look like those given for the partial sums of another series in A128507.

Rationals (partial sums) Theta(3,n) := Sum_{j=1..n} 1/(2*j-1)^3 (in lowest terms). The limit of these rationals is Theta(3) = (1-1/2^3)*Zeta(3) approximately 1.051799790 (Zeta(n) is the Euler-Riemann zeta function).

This is a member of the k-family of rational sequences Theta(k,n) := Sum_{j=1..n} 1/(2*j-1)^k, k >= 1, which coincides for k=1 with A025550/A025547 (but only for the first 38 terms), for k=2 with A120268/A128492, for k=3 with a(n)/A128507(n) (the denominators may depart for higher n values), A120269/A128493 and A164656/A164657, for k=4 and 5, respectively.