Harmonic Number Sum Congruence
Leroy Quet
qq-quet at mindspring.com
Fri Jan 9 04:55:47 CET 2004
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
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