sum Fibonacci(m+1-k) /k

[Leroy Quet]

I posted the below to sci.math and rec.puzzles as a to-be-proved puzzle 
[if my result is even  true].
I only post it to seq.fan, though, because the integer sequence

m! * sum{k=1 to m}  F(m+1-k) /k, 

1, 3, 17, 110,...

does not seem to be in the EIS, and might have some other interesting 
properties.

thanks,
Leroy Quet

---
 

This might be easy, but try to prove:


For m = EVEN positive integer,


(m+1) always divides


m! * sum{k=1 to m}  F(m+1-k) /k ,


where F(n) is the n_th Fibonacci number.
(F(1) = 1, F(2) = 1, F(m+2) = F(m+1) +F(m))






And in general, for more advanced players...:

Let H(0,m) = 1/m,

for all positive integers m.

Let H(n,m) = sum{k=1 to m} H(n-1,k) ,

for all positive integers n.



Then, for all EVEN positive integers m, 
and for all nonnegative integers n, 


(m+2n+1) always divides


m! (sum{k=1 to m}  F(m+1-k) * H(n,k) ).




For example,


(m+3) always divides, for even m,


m! *sum{k=1 to m} F(m+1-k) H(k),


where H(k) = 1 +1/2 +1/3 +..+ 1/k,
the kth harmonic number.


thanks,
Leroy 
     Quet