# Harmonic Number Sum Congruence

Leroy Quet qq-quet at mindspring.com
Wed Jan 14 23:23:08 CET 2004

```I forgot to post the following to seq.fan:

By the way, the sequence of

{m! h(m) (mod {m+1})}  -> 1, 1 ,0, 3, 4, 2,.. (?)

does not seem to be in the EIS either.

Is there a closed form for this sequence? ...

(...which also equals
{sum{k=1 to m}  H(k) k! (m-k)! (mod {m+1})})

Regarding:
>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

```