[Roland.Bacher at ujf-grenoble.fr: Bernoulli numbers and arithmetic progressions]: Corrections
Roland Bacher
Roland.Bacher at ujf-grenoble.fr
Wed Feb 4 14:35:54 CET 2004
Here a few precisions and additional informations (I hope
with not too many errors)
B(2n) (if my memory is correct) is always a rational whose
denominator is well understood (by van Staudt's Theorem).
Its numerator is more mysterious. A prime involved in the
numerator of B(2n) is called exceptional if it does not
divide n. Cyclotomic field associated
to exceptional primes have special properties
(related, I believe, with partial proofs of Fermat's
last theorem which fail for such primes).
A little experimentation with exceptional primes (37,59,67,103)
shows that the exceptional indices 2n for which p divides
the numerator of B(2n) form seemingly
arithmetic progressions of step (p-1)/2. The conjecture
stated below follows then probably easily. (But the information
concerning the starting points, respectively the set of
exceptional primes, remain mysterious).
I don't know if such results are known, has anybody a reference?
For checking such claims it is perhaps best to work only in the field
of p-adic numbers (where p is the exceptional prime which is studied).
There is a linear (in B(k)'s) recurrence relation for B(2n)
(involving binomial
coefficients). Clearing denominators, one can then work over
Z/(p^a Z) with a fairly small which makes computations possible perhaps
up to 10^8 or so (I have to think this over a little more before
beeing able to implement it properly, perhaps in a C++ program).
Roland Bacher
Here an amusing remark using the data below:
all values yielding 37 are of the form: 574+666*k, k=0,1,2,3,4,...
and form thus an arithmetic progression with step 666=18*37=((37-1)/2)*37;
all values yielding 59 are of the form: 1269+1711*k, k=0,1,2,3
and 1711=28*59=((59-1)/2)*59
the two values yielding 67 are at distance 2211=((67-1)/2)*67
Conjecture: all indices yielding a given prime p form an arithmetic
progression of step ((p-1)/2)*p.
Is this obvious?
If yes, the only important data would be the beginning of the
arithmetic progression for a given prime.
By the way, in order to check this conjecture experimentally in
a few more cases it is enough to compute the corresponding
p-valuations. This can be done using (for instance) a recursive
formula for Bernoulli numbers and working in the finite ring
Z/(p^a Z) with a>0 big enough (in any case, a such that p^a>2n
should work). This trick should allow to push computations quite a lot
further.
Roland Bacher
Hi, Neal
GP/PARI CALCULATOR Version 2.2.8 (development CHANGES-1.887)
i686 running cygwin (ix86 kernel) 32-bit version
compiled: Jan 13 2004, gcc-3.3.1 (cygming special)
(readline v4.3 enabled, extended help not available)
system pIV 2.533ghz 2 gig ram xp pro
Here are some more terms in Pari. You will have to allocatemen() for m1 or
m2 > 7000
I had the set for below m1=1,m2 = 10000 but knew it would crash for more
memory so
I stopped. BTW, you don't need arrays for this sequence. Some speed gain is
realized also
by setting 2*n to n2 befor the bernfrac() call.
\\ c = start index of next term not computed,m1 = start, me=end
bern2(c,m1,m2) =
{
for(n=m1,m2,
n2=n+n;
a = numerator(bernfrac(n2)/(n2)); \\ A001067
b = numerator(bernfrac(n2)/(n2*(n2-1))); \\ A046968
if(a <> b,print("A("c") = "n","a/b);c++)
)
}
1,574,37
2,1185,103
3,1240,37
4,1269,59
5,1376,131
6,1906,37
7,1910,67
8,2572,37
9,2689,283
10,2980,59
11,3238,37
12,3384,101
13,3801,691
14,3904,37
15,4121,67
16,4570,37
17,4691,59
18,4789,157
19,5236,37
system pIII 1 ghz 256k ram xp pro
This broke down at 4570 giving a huge number as the result. must be memory
pr p3 /pari bug.
2689,4691,4789 are primes.
Looks like you have 1 more yet one more sequence.
Cino
>From: "N. J. A. Sloane" <njas at research.att.com>
>Reply-To: njas at research.att.com
>To: eclark at math.usf.edu, somos at grail.cba.csuohio.edu, seqfan at ext.jussieu.fr
>Subject: Re Bernoulli numbers
>Date: Tue, 3 Feb 2004 15:39:10 -0500 (EST)
>
>Thanks, Edwin!
>
>%I A090495
>%S A090495 574,1185,1240,1269,1376,1906,1910
>%N A090495 Numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different
>from numerator(Bernoulli(2*n)/(2*n*(2*n-1))).
>%C A090495 n such that A001067 is different from A046968.
>%O A090495 1,1
>%K A090495 nonn,more
>%Y A090495 Cf. A090496, A001067, A046968.
>%p A090495 a:=n->numer(bernoulli(2*n)/(2*n)):
>b:=n->numer(bernoulli(2*n)/(2*n*(2*n-1))): for n from 1 to 2000 do if
>a(n)<>b(n) then print(n,a(n)/b(n)); fi; od:
>%t A090495 a[n_] := Numerator[BernoulliB[2n]/(2n)] (* A001067 *); b[n_] :=
>Numerator[BernoulliB[2n]/(2n(2n-1))] (* A046968 *); For[n=1, n <= 580, n++,
>If[ a[n] != b[n], Print[n, " ", a[n]/b[n]] ] ]
>%A A090495 njas, Feb 03 2004
>%E A090495 a(1) discovered by Michael Somos, Feb 01 2004. a(2)-a(7) from
>Edwin Clark, Feb 03, 2004.
>
>%I A090496
>%S A090496 37,103,37,59,131,37,67
>%N A090496 Ratio of numerator(Bernoulli(2*n)/(2*n)) to
>numerator(Bernoulli(2*n)/(2*n*(2*n-1))) for n's for which they are
>different.
>%C A090496 A001067(n) / A046968(n) when they are different.
>%O A090496 1,1
>%K A090496 nonn,more
>%Y A090496 Cf. A090496, A001067, A046968.
>%A A090496 njas, Feb 03 2004
>%E A090496 a(1) discovered by Michael Somos, Feb 01 2004. a(2)-a(7) from
>Edwin Clark, Feb 03, 2004.
>
>NJAS
_________________________________________________________________
Scope out the new MSN Plus Internet Software optimizes dial-up to the max!
http://join.msn.com/?pgmarket=en-us&page=byoa/plus&ST=1
----- End forwarded message -----
----- End forwarded message -----
More information about the SeqFan
mailing list