Harmonic Number Sum Congruence

[Leroy Quet]

I wrote:

>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}.)
>




We can rewrite the above so it becomes, perhaps to some reading this, 
more simply stated and more natural.

(Take sum from 1 to m, instead of from 1 to m-1)


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



is congruent to



m! *h(m)   (mod(m+1)),


where h(m) = sum{k=0 to floor((m-1)/2)}  1/(m -2k)  .



Now, h(m) =
sum{k=0 to floor((m-1)/2)}  1/(m -2k)
is interesting in itself, for it is (analogously to "double-factorials") 
the sum of the reciprocal of EVERY-OTHER positive integer <=m 
(even or odd, depending on m).


Neither sequence:

m!*h(m) 
nor
m!!*h(m)

nor
sum{k=1 to m}  H(k) k! (m-k)!

seems to be in the EIS either, if I calculated correctly by hand.

Thanks,
Leroy
 Quet

>
>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