cp's OEIS Frontend

Numerator of Sum_{k=0..n-1} 1/((k+1)(2k+1)) (denominator is A111876). - Paul Barry, Aug 19 2005

Numerator of the sum of all matrix elements of n X n Hilbert matrix M(i,j) = 1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk, Apr 11 2006

Numerator of the 2n-th alternating harmonic number H'(2n) = Sum ((-1)^(k+1)/k, k=1..2n). H'(2n) = H(2n) - H(n), where H(n) = Sum_{k=1..n} 1/k is the n-th Harmonic Number. - Alexander Adamchuk, Apr 11 2006

a(n) almost always equals A117731(n) = numerator(n*Sum_{k=1..n} 1/(n+k)) = numerator(Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1)) but differs for n = 14, 53, 98, 105, 111, 114, 119, 164. - Alexander Adamchuk, Jul 16 2006

Sum_{k=1..n} 1/(n+k) = n!^2 *Sum_{j=1..n} (-1)^(j+1) /((n+j)!(n-j)!j). - Leroy Quet, May 20 2007

Seems to be the denominator of the harmonic mean of the first n hexagonal numbers. - Colin Barker, Nov 19 2014

Numerator of 2*n*binomial(2*n,n)*Sum_{k = 0..n-1} (-1)^k* binomial(n-1,k)/(n+k+1)^2. Cf. A049281. - Peter Bala, Feb 21 2017

From Peter Bala, Feb 16 2022: (Start)

2*Sum_{k = 1..n} 1/(n+k) = 1 + 1/(1*2)*(n-1)/(n+1) - 1/(2*3)*(n-1)*(n-2)/((n+1)*(n+2)) + 1/(3*4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) - 1/(4*5)*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) + - .... Cf. A101028.

2*Sum_{k = 1..n} 1/(n+k) = n - (1 + 1/2^2)*n*(n-1)/(n+1) + (1/2^2 + 1/3^2)*n*(n-1)*(n-2)/((n+1)*(n+2)) - (1/3^2 + 1/4^2)*n*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + (1/4^2 + 1/5^2)*n*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) - + .... Cf. A007406 and A120778.

These identities allow us to extend the definition of Sum_{k = 1..n} 1/(n+k) to non-integral values of n. (End)

a(n) almost always equals A082687(n), but differs for n in A125740.

p divides a((p-1)/3) for primes p in A002476, that is, primes of form 6*n + 1. - Alexander Adamchuk, Jul 16 2006

Inverse product of all matrix elements of n X n Hilbert matrix M(i,j) = 1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk, Apr 12 2006

The n X n matrix with A(i,j) = 1/(i+j-1)! (i,j = 1..n) has determinant (-1)^floor(n/2)/a(n). - Mikhail Lavrov, Nov 01 2022

Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) = A117731(n) / A117664(n) = 2n * H'(2n) = 2n * A058313(2n) / A058312(2n), where H'(2n) is 2n-th alternating sign Harmonic Number. H'(2n) = H(2n) - H(n), where H(n) is n-th Harmonic Number. - Alexander Adamchuk, Apr 23 2006

Solves the following arithmetic problem. Let (3) q#(i/j)=(q+h) be defined thus: Given q, an unknown fraction, q=qd/d, of integer value (initially) and h equal to any desired integer value and j equal to any desired integer value and i equal to j(q+h)+h.

Let '#' denote the repeated addition of the denominator j to the denominator of q and the repeated addition of the numerator i to the numerator of q, each addition recursively replacing the prior q fractions numerator and denominator respectively. This results in a series of fractions of mostly non-integer values.

After the first three iterations for any case would result in the following fractions: 1) (qd+i) / (d+j) 2) ((qd+i)+i) / ((d+j)+j) 3) (((qd+i)+i)+i) / (((d+j)+j)+j)

The question is, what is the smallest initial denominator, d (of q=qd/d) that in the course of the repeated additions will result in fractions of integer value, where every integer, from the initial (q+1)...(q+h) will be formed?

For example, let q = 4, h = 5, so (q+h)=9 and j = 1, so that for i we have i = 14. So in terms of q#(i/j)=(q+h) we have 4#(14/1)=9.

In this sample, since h=5 and j=1 we get from (2) above that d(5,1) = 252 as the solution. Applying this solution, we see then that the initial numerator of q, or q*d, becomes 4*252 = 1008 and the initial denominator is 252. Alternatively, in terms of q#(i/j)=(q+h) we have (1008/252)#(14/1)=9. We see that the repeated additions yield:

1) 1008 +14 and 252 +1 ~ 1022/253

2) 1022 +14 and 253 +1 ~ 1036/254

...

27) 1372 +14 and 277 +1 ~ 1386/279

28) 1386 +14 and 278 +1 ~ 1400/280 =5

...

63) 1876 +14 and 314 +1 ~ 1890/315 =6

...

108) 2506 +14 and 359 +1 ~ 2520/360 =7

...

168) 3346 +14 and 419 +1 ~ 3360/420 =8

...

252) 4522 +14 and 503 +1 ~ 4536/504 =9

Note that h=5 and j=1 were chosen so that d(5,1) = a(5) of the sequence. Also note that by the definition of q#(i/j)=(q+h) all answers are only a function of h and j.

Note also that q=qd/d can also be expressed as 0=0d/d and that any q#(i/j)=(q+h) can be expressed as 0#(i/j)=h [after adjustments are made to i]. For instance 4#(14/1)=9 is the equivalent of 0#(10/1)=5 in terms of d(h,j).

Interestingly: For any q#(i/j)=p and r#(s/j)=t then (q+r)#((i+s)/j)=(p+t). Also for any q#(i/j)=p then (qr)#((ir)/j)=(pr).

The sequence A025558 can be calculated from this formula when h = j, in otherwords using the sequence of d(1, 1)...d(n, n). i.e. a(7) = 735 = d(7,7) = lcm(49,21,35,7,21,7,1) a(8) = 2240 = d(8,8) = lcm(64,28,16,10,32,416,1)

Denominator of the sum of all elements of n X n Hilbert Matrix M[i,j] with alternate signs. M[i,j] = 1/(i+j-1)(i,j = 1..n). - Alexander Adamchuk, Apr 11 2006

Nome q(m) = x exp(8 * (Sum_{n>0} a(n) * x^n / n!) / (Sum_{n>=0} binomial(2n, n)^2 * x^n)) where x = m / 16.

The Fricke reference on page 2 has equation "(3) Pi i omega = -Pi K'/K = log k^2 - 4 log 2 + F_1(1/2, 1/2; k^2) / F(1/2, 1/2, 1; k^2), wo F_1 und F ..." where F_1 = 8 * Sum_{n>0} a(n) * x^n / n! with x = m / 16 = (k / 4)^2. - Michael Somos, Jul 14 2013

The numerators are given in A101028.

One third of the denominator of the finite differences of the series of sums of all matrix elements of n X n Hilbert matrix M(i,j)=1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk, Apr 11 2006