Ugly But Interesting Harmonic Number Sequence

[Leroy Quet]

First, thanks to Max A. and Richard Mathar for their replies to this 
thread.


The result is easily derived from the earlier result of mine:
(View with fixed-width font.)

 n
---
\   (H(2k) - H(n-k)) (2k)!
/   ----------------------- = 0,
---      k! (k+n+1)!
k=0

for all nonnegative integers n.

(H(m) = sum{j=1 to m} 1/j, the mth harmonic number. H(0) = 0.)

It has been some time since I found the above result. And, since I cannot 
at this time reverse-engineer a proof, I am not ABSOLUTELY certain that 
it is correct.

In any case, I just submitted the following:

>%I A124235
>%S A124235 1,1,17,877
>%N A124235 a(n) = numerator of (sum{k=1 to n} H(2k)(2k)!/(k!(k+n+1)!) = 
>sum{k=0 to n-1} H(n-k)(2k)!/(k!(k+n+1)!)), where H(k) = sum{j=1 to k} 1/j 
>(ie the kth harmonic number).
>%Y A124235 A124236
>%O A124235 0
>%K A124235 ,frac,more,nonn,
>%A A124235 Leroy Quet (qq-quet at mindspring.com), Oct 22 2006

>%I A124236
>%S A124236 2,3,144,30240
>%N A124236 a(n) = denominator of (sum{k=1 to n} H(2k)(2k)!/(k!(k+n+1)!) = 
>sum{k=0 to n-1} H(n-k)(2k)!/(k!(k+n+1)!)), where H(k) = sum{j=1 to k} 1/j 
>(ie the kth harmonic number).
>%Y A124236 A124235
>%O A124236 1
>%K A124236 ,frac,more,nonn,
>%A A124236 Leroy Quet (qq-quet at mindspring.com), Oct 22 2006

My request is to whomever extends the sequences: Once the sequences 
appear in the database, could you please try the values for both sums, 
just as a check that the sums are the same for the first N values of n at 
least?

thanks,
Leroy Quet

PS: There are a few seq.fan people who could EASILY prove that the two 
sums equal each other, I bet.