Bernoulli numbers and arithmetic progressions

[cino hilliard]

Hi,

>From: Hans Havermann <hahaj at rogers.com>
>To: seqfan at ext.jussieu.fr
>Subject: Re: Bernoulli numbers and arithmetic progressions
>Date: Sun, 8 Feb 2004 22:27:53 -0500
>
>I asked:
>
>>What are the values of p*k-(p-1)/2 where Mod[Numerator[BernoulliB[2k]], p] 
>>== 0 for p=37?
>>Every third term in this calculated sequence, it seems, is a number that 
>>does not figure in the *actual* 37-appearance sequence.
>I was still convinced however, ignoring every third term, that 37 appears 
>periodically with a difference of 666 between adjacent terms. As I 
>calculated and collated the differences between adjacent terms of a range 
>of irregular primes, hoping to differentiate the clearly periodic ones from 
>the aberrations, I ran into a rude surprise as I upped my values of k. Same 
>calculation as above, ignoring every third term, the differences between 
>adjacent terms for p=37 are:
>
>{666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 
>666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 666, 
>666, 1332, 666, 666, 666, 1332, 111, 1221, 148, 1184, 185, 1147, 222, 1110, 
>259, 1073, 296, 1036, 333, 999, 370, 962, 407, ...}
>
>I feel a little discombobulated and would appreciate someone
New word you just bobbed in? :-)
>confirming or denying this progression.

This progression fails for me too with pari but i get different numbers.
Some excerpts  with k, n, constant. It appears as p increases the failure is 
further along.
37 fails around 800, 59 around 2000 and 67 around 2400. I think the problem 
is the
we have a different problem. The org algorithm involving

          a = numerator(bernfrac(n2)/(n2));           \\ A001067
          b = numerator(bernfrac(n2)/(n2*(n2-1)));    \\ A046968

required a and b  be different. Your every third step may violate this at 
some point. Try pari
below. You will see the failure begins after the (p-1)/2 th entry. 
Therefore, you can form your
conjecture: every third entry up to (p-1)/2.

Also inspection shows n = ((2k+1)p+1)/2 for a given irr prime p and some k. 
I was able to
find the index for irr prime 257, n= 20946 and 293,  n = 22708 in my spread 
sheet. So far
The search for n for 311 is in progress. I have computed to A(68) to date.

http://groups.yahoo.com/group/B2LCC/files/Bernoulli/


(00:06) gp > bern4(800,37)
16 574 666
37 1351 666
70 2572 666
...
407 15041 666
430 15892 666
111 = first fail 666
18  =  number correct


(23:55) gp > bern4(2000,59)
22 1269 1711
59 3452 1711
...
1066 62865 1711
1121 66110 1711
472 = first fail from 1711
29   = number correct

(00:41) gp > bern4(2400,67)
29 1910 2211
67 4456 2211
128 8543 2211
...
1349 90350 2211
1407 94236 2211
1448 96983 2211
472 = first fail from 2211
29   = number correct

bern4(n,p) =
          {
           c=0;
           c2=0;
           c3=0;
           a=vector(n);
           a2=vector(n);
           ak=vector(n);
           for(k=1,n,
           if(numerator(bernfrac(k+k))%p==0,
           c++;
           a[c]=p*k-(p-1)/2;
           ak[c]=k;
            )
            );
           for(x=1,c,
           if(x%3,c2++;a2[c2]=a[x])
            );
           d=a2[2]-a2[1];
           forstep(x=1,c,2,
           d2 =a2[x+1]-a2[x];
           if(a2[x+1]>0 && d==d2,c3++;print(ak[x]" "a[x]" "d2),
           f=a2[x+1]- a2[x];
           break)
                  );
           print( f" = first fai from "d);
           print(c3" = number correct")
           }

Have fun,
Cino
FB

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