Harmonic Number Sum Congruence

[Leroy Quet]

Below is some of a post I made recently to sci.math.

I have added some more seq.fan-friendly material and edited the rest of 
my original (truncated) message.




Let H(n) = sum{k=1 to n} 1/k, 
the n_th harmonic number.

I suppose that, for each m = positive integer,



sum{k=1 to m-1}  H(k) k! (m-k)!

is congruent to

(1/2) m! H(floor(m/2))) (-1)^m 

(mod (m+1)) .


(I do not know if the result is trivial {or am certain it is correct}.)


I decided to post this message to seq.fan because the sequence formed from
sum{k=1 to m-1}  H(k) k! (m-k)!
is not in the database.

It would be more natural too take the sum to m instead of (m-1).
So, the sequence
sum{k=1 to m}  H(k) k! (m-k)!
begins (figured by hand):

1, 4, 16, 73, 388, ...

And, no, this sequence does not seem to be in the EIS either.



thanks,
Leroy Quet