On Bernoulli numbers

[benoit]

Bonjour,

Let N(n,m)=r(m)*(m^(2n)-1)*B(2n) where m is any fixed integer >=1, 
where B(2n) is the 2n-th Bernoulli number and where r(m) is a rational 
value such that N(n,m) is an integer value for all n>0. r(m) always 
exist since r(m)=m works. I'm looking for the least rational value 
A(m)=r(m) such that for fixed m, N(n, m) is an integer value for  all 
n>0.  If m=2, A(2)=2 and N(n,2)=2*(4^n-1)*B(2n) are the Genocchi 
numbers of first kind.

I get (for m>=3 that's experimental) :

A(1)=1 ; A(2)=2 ; A(3)=3/4 ; A(4)=2 ; A(5)=5/4 ; A(6)=6 ; A(7)=7/8 ; 
A(8)=2 ; A(9)=3/8 ; A(10)=10/3 ; A(11)=11/4 ; A(12)=6 ; A(13)=13/4 ; 
A(14)=14 ;
A(15)=15/16 ; A(16)=2 ; A(17)=17/48 ; A(18)=6 ; A(19)=19/12 ; A(20)=10 
; A(21)=21/4 ; A(22)=22 ; A(23)=23/8 ; A(24)=6/5 ; A(25)=5/8 
;A(26)=26/45 ;
A(27)=3/4 ; A(28)=14/9 ; A(29)=29/4 ; A(30)=30 ; A(31)=31/32 ; A(32)=2 
; A(33)=33/32 ; A(34)=34 ; A(35)=35/12

It appears that if m is a power of 2, A(m)=2 and more generally, 
letting A(m)=P(m)/Q(m)  (P(m),Q(m))=1 it seems that :

  Sequence P(m) := 1,2,3,2,5,6,7,2,3,10,11,6,13,14,15,2,17,6... = 
largest squarefree number dividing m ( A007947(m) )

I found no general rule for sequence Q(m) := 
1,1,4,1,4,1,8,1,8,3,4,1,4,1,16,1,48,1,12,1,4,1,8,5,8,....

Can anyone confirm/extend  sequence (A(m)) m>=1 ? Is there any simple 
rule for sequence (Q(m)) m>=1 ?

Thanks
Benoit Cloitre







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