On the divisibility of numerator of B(2n) by primes

[benoit]

Following TD Noe. For a given prime p here is an essay to define  
general rules for the sequence of n such that p divides the numerator  
of B(2n). I considered whether p is irregular or not.


(i) if q is a regular prime :

For a regular odd prime q, the sequence of n such that q divides the  
numerator of  B(2n) "is" given by :

{q*k}_{k>=0 and k not congruent to (q-1)/2}

Should be a direct consequence of Von Staudt and Clausen.

Exemples :

(i.1) the sequence of n such that 5 divides the numerator of  B(2n)  
(divided by 5) "is" given by :

1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51 
,53,55,57,....

and even numbers = multiple of (5-1)/2=2 are missing.

(i.2) the sequence of n such that 17 divide the numerator of  B(2n)  
(divided by 17) "is" given by :

1,2,3,4,5,6,7,9,10,11,12,13,14,15,17,...

and multiple of (17-1)/2=8 are missing.


(ii) if p is an irregular prime :

  For a given irregular prime p let  {x1,x2,...,xt} denotes the set of  
distinct values 1<=x1<...<xt<(p-1)/2 such that p divides numerator of  
B(2*xi) for i=1,...,t.

Then I suspect the sequence of n such that p divides the numerator of  
B(2n) "to be" given by :

{p*k} U_{i=1}^{t} {(p-1)/2*k+xi}   for k>=0.

Exemples :

(ii.1) the sequence of n such that 37 divides the numerator of B(2n) is  
(here there is a single value x1=16) :

16,34,37,52,70,74,88,106,111,124,142,148,160,178,185,196,214,222,232,250 
,259,268,286,

which "is" the sequence {37*k} U {18*k+16} k>=0 .

(ii.2) the sequence of n such that 157 divides the numerator of  B(2n)  
is (here there are 2 values x1=31 and x2=55)  :

31,55,109,133,157,187,211,265,289,314,343,367,

which "is" the sequence {157*k} U {78*k+31} U {78*k+55} k>=0 .


If all of this is true :  B(2n)/(2n)/(2n-1) has a different numerator  
than  B(2n)/(2n) means that at least one prime factor p of (2n-1)  
divides B(2n).

Since (2n,2n-1)=1 clearly p can't be regular from (i). Otherwise  
2n=2*p*k contradicting (2n,2n-1) =1.

Determining the sequence of n for which numerator of B(2n)/(2n)/(2n-1)  
differs from numerator of  B(2n)/(2n) should imply the computation of  
{(xi)} for all irregular primes.

For info. I submitted this complement of "indexed" sequences for  
irregular primes (as A073277) :


%S A000001  
0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,1,0,2, 
0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,0,1,0,0,1,1,1,1,0,0,0,0,1,0,2,0,0, 
0,2,0,1,0,1,1,0,1,0,1,0,0,0,0,1,1,2,0,0,3,0,0,0
%N A000001 number of k, 1<=k<=(n-1)/2, such that n divides the  
numerator of B(2k) where n runs through the odd primes.
%F A000001 a(n) = 0 if n is a regular prime ; a(n) > 0 if n is an  
irregular prime


B.Cloitre 
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: text/enriched
Size: 2876 bytes
Desc: not available
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20040210/21e4525f/attachment.bin>