Ugly But Interesting Harmonic Number Sequence

[Max A.]

On 10/21/06, Leroy Quet <qq-quet at mindspring.com> wrote:
> Let H(n) = sum{k=1 to n} 1/k, the nth harmonic number.
> (H(0) = 0.)
>
> Then, for n = any *positive* integer,
>
> a(n) = ( (2n)!*(1 -H(2n))/2 + (2n+1)!*(H(n+1)/(n+2) - H(2n+2)/(2n+3))
> )/(n+1)
>
> is always an integer.
>
>
> I based this result on an identity I found years ago which I cannot
> remember how I got it. So perhaps the above result is wrong.

It is correct.

First off, I will prove that the numerator of a(n), i.e.,
(2n)!*(1-H(2n))/2 + (2n+1)!*(H(n+1)/(n+2) - H(2n+2)/(2n+3)), is
integer.
For n=1, it simply equals 0.
For n>1, it is easy to see that the terms (2n)!*(1-H(2n))/2 and
(2n+1)!*H(n+1)/(n+2) are integer. To finish the proof we need to show
that (2n+1)!*H(2n+2)/(2n+3) is integer.

Note that (2n+1)!*H(2n+2) = (2n+1)!*H(2n+1) + (2n+1)!/(2(n+1)) is
integer. Also note that
(2n+1)!*H(2n+2)/(2n+3) = -s(2n+3,2)/(2n+2)/(2n+3) where s(2n+3,2) is
Stirling number of the first kind, see
http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html
Since gcd((2n+2),(2n+3))=1, it is enough to show that the number
(2n+2)!*H(2n+2)/(2n+3) = -s(2n+3,2)/(2n+3) is integer, i.e., (2n+3)
divides s(2n+3,2).
We will prove that for odd m>0, m divides s(m,2). We have (see formula
(8) at MathWorld)
s(m,1)*m^1 + s(m,2)*m^2 + ... + s(m,m)*m^m = m!
Hence, m divides s(m,2) as soon as m divides (m! - s(m,1)*m)/m^2. But
(m! - s(m,1)*m)=0 for odd m, proving that m divides s(m,2). Therefore,
(2n+3) divides s(2n+3,2) and (2n+1)!*H(2n+2)/(2n+3) is integer.

To prove that a(n) is integer, we will show that its numerator is 0
modulo n+1. We have
(2n)!*(1-H(2n))/2 = -(2n)!/(2(n+1)) (mod n+1)
and
(2n+1)!*(H(n+1)/(n+2) - H(2n+2)/(2n+3))
= (2n+1)!*(1/((n+1)*(n+2)) - (1/(n+1)+1/(2(n+1)))/(2n+3))
= (2n+1)!*n/(2*(n+1)*(n+2)*(2n+3)) (mod n+1)
Hence,
-(2n)!/(2(n+1)) + (2n+1)!*n/(2*(n+1)*(n+2)*(2n+3))
= -3*(2n)!/((n+2)*(2n+3)) = 0 (mod n+1)
completing the proof that a(n) is integer.

Max