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<DIV><FONT size=1></FONT>i found that 6568, 37 is true pehaps is a(22)</DIV>
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<DIV><BR>>>> Roland Bacher <Roland.Bacher@ujf-grenoble.fr> 02/04
2:35 >>><BR>Here a few precisions and additional informations (I
hope<BR>with not too many errors)<BR><BR>B(2n) (if my memory is correct) is
always a rational whose<BR>denominator is well understood (by van Staudt's
Theorem).<BR>Its numerator is more mysterious. A prime involved in
the<BR>numerator of B(2n) is called exceptional if it does not<BR>divide n.
Cyclotomic field associated<BR>to exceptional primes have special properties
<BR>(related, I believe, with partial proofs of Fermat's <BR>last theorem which
fail for such primes).<BR><BR>A little experimentation with exceptional primes
(37,59,67,103)<BR>shows that the exceptional indices 2n for which p
divides<BR>the numerator of B(2n) form seemingly <BR>arithmetic progressions of
step (p-1)/2. The conjecture <BR>stated below follows then probably easily. (But
the information<BR>concerning the starting points, respectively the set of
<BR>exceptional primes, remain mysterious).<BR><BR>I don't know if such results
are known, has anybody a reference?<BR><BR>For checking such claims it is
perhaps best to work only in the field <BR>of p-adic numbers (where p is the
exceptional prime which is studied).<BR>There is a linear (in B(k)'s) recurrence
relation for B(2n) <BR>(involving binomial <BR>coefficients). Clearing
denominators, one can then work over<BR>Z/(p^a Z) with a fairly small which
makes computations possible perhaps <BR>up to 10^8 or so (I have to think this
over a little more before<BR>beeing able to implement it properly, perhaps in a
C++ program).<BR><BR>Roland Bacher<BR><BR><BR>Here an amusing remark using the
data below:<BR><BR>all values yielding 37 are of the form: 574+666*k,
k=0,1,2,3,4,...<BR>and form thus an arithmetic progression with step
666=18*37=((37-1)/2)*37;<BR><BR>all values yielding 59 are of the form:
1269+1711*k, k=0,1,2,3<BR>and 1711=28*59=((59-1)/2)*59<BR><BR>the two values
yielding 67 are at distance 2211=((67-1)/2)*67<BR><BR>Conjecture: all indices
yielding a given prime p form an arithmetic<BR>progression of step
((p-1)/2)*p.<BR><BR>Is this obvious?<BR><BR>If yes, the only important data
would be the beginning of the <BR>arithmetic progression for a given
prime.<BR><BR>By the way, in order to check this conjecture experimentally
in<BR>a few more cases it is enough to compute the corresponding
<BR>p-valuations. This can be done using (for instance) a recursive <BR>formula
for Bernoulli numbers and working in the finite ring <BR>Z/(p^a Z) with a>0
big enough (in any case, a such that p^a>2n<BR>should work). This trick
should allow to push computations quite a lot<BR>further.<BR><BR>Roland
Bacher<BR><BR><BR>Hi, Neal<BR>GP/PARI CALCULATOR Version 2.2.8 (development
CHANGES-1.887)<BR> i686 running cygwin (ix86
kernel) 32-bit version<BR> compiled: Jan 13 2004,
gcc-3.3.1 (cygming special)<BR> (readline v4.3 enabled,
extended help not available)<BR><BR>system pIV 2.533ghz 2 gig ram xp
pro<BR><BR>Here are some more terms in Pari. You will have to allocatemen() for
m1 or <BR>m2 > 7000<BR>I had the set for below m1=1,m2 = 10000 but knew it
would crash for more <BR>memory so<BR>I stopped. BTW, you don't need
arrays for this sequence. Some speed gain is <BR>realized also<BR>by setting 2*n
to n2 befor the bernfrac() call.<BR><BR>\\ c = start index of next term not
computed,m1 = start, me=end<BR>bern2(c,m1,m2)
=<BR> {<BR>
for(n=m1,m2,<BR>
n2=n+n;<BR> a =
numerator(bernfrac(n2)/(n2));
\\ A001067<BR> b =
numerator(bernfrac(n2)/(n2*(n2-1))); \\
A046968<BR> if(a <>
b,print("A("c") =
"n","a/b);c++)<BR>
)<BR>}<BR><BR>1,574,37<BR>2,1185,103<BR>3,1240,37<BR>4,1269,59<BR>5,1376,131<BR>6,1906,37<BR>7,1910,67<BR>8,2572,37<BR>9,2689,283<BR>10,2980,59<BR>11,3238,37<BR>12,3384,101<BR>13,3801,691<BR>14,3904,37<BR>15,4121,67<BR>16,4570,37<BR>17,4691,59<BR>18,4789,157<BR>19,5236,37<BR><BR>system
pIII 1 ghz 256k ram xp pro<BR>This broke down at 4570 giving a huge number as
the result. must be memory <BR>pr p3 /pari bug.<BR>2689,4691,4789 are
primes.<BR><BR>Looks like you have 1 more yet one more
sequence.<BR><BR>Cino<BR><BR><BR><BR><BR>>From: "N. J. A. Sloane"
<njas@research.att.com><BR>>Reply-To: njas@research.att.com<BR>>To:
eclark@math.usf.edu, somos@grail.cba.csuohio.edu,
seqfan@ext.jussieu.fr<BR>>Subject: Re Bernoulli numbers<BR>>Date: Tue, 3
Feb 2004 15:39:10 -0500 (EST)<BR>><BR>>Thanks, Edwin!<BR>><BR>>%I
A090495<BR>>%S A090495 574,1185,1240,1269,1376,1906,1910<BR>>%N A090495
Numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different <BR>>from
numerator(Bernoulli(2*n)/(2*n*(2*n-1))).<BR>>%C A090495 n such that A001067
is different from A046968.<BR>>%O A090495 1,1<BR>>%K A090495
nonn,more<BR>>%Y A090495 Cf. A090496, A001067, A046968.<BR>>%p A090495
a:=n->numer(bernoulli(2*n)/(2*n)):
<BR>>b:=n->numer(bernoulli(2*n)/(2*n*(2*n-1))): for n from 1 to 2000 do if
<BR>>a(n)<>b(n) then print(n,a(n)/b(n)); fi; od:<BR>>%t
A090495 a[n_] := Numerator[BernoulliB[2n]/(2n)] (* A001067 *); b[n_] :=
<BR>>Numerator[BernoulliB[2n]/(2n(2n-1))] (* A046968 *); For[n=1, n <=
580, n++, <BR>>If[ a[n] != b[n], Print[n, " ", a[n]/b[n]] ] ]<BR>>%A
A090495 njas, Feb 03 2004<BR>>%E A090495 a(1) discovered by Michael Somos,
Feb 01 2004. a(2)-a(7) from <BR>>Edwin Clark, Feb 03, 2004.<BR>><BR>>%I
A090496<BR>>%S A090496 37,103,37,59,131,37,67<BR>>%N A090496 Ratio of
numerator(Bernoulli(2*n)/(2*n)) to
<BR>>numerator(Bernoulli(2*n)/(2*n*(2*n-1))) for n's for which they are
<BR>>different.<BR>>%C A090496 A001067(n) / A046968(n) when they are
different.<BR>>%O A090496 1,1<BR>>%K A090496 nonn,more<BR>>%Y A090496
Cf. A090496, A001067, A046968.<BR>>%A A090496 njas, Feb 03 2004<BR>>%E
A090496 a(1) discovered by Michael Somos, Feb 01 2004. a(2)-a(7) from
<BR>>Edwin Clark, Feb 03,
2004.<BR>><BR>>NJAS<BR><BR>_________________________________________________________________<BR>Scope
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