(2^n+1)/3

[Gottfried Helms]

Am 20.05.05 08:50 schrieb kohmoto:


>>
>>     Hello, Seqfans
>>     Once I realized that if M_n is a Mersenne prime then (M_n+2)/3 is
>> also prime. 2<n
>>     And I knew that it is called "Bateman and Shefridge and Wagstaff's
>> conjecture ".
>>     Does anyone know the exact description of it?
>>
>>     %I A000001
>>     %S A000001 3, 11, 43, 2731, 43691
>>     %N A000001 (M_n+2)/3. Where M_n is Mersenne prime 2^n-1.
>>     %C A000001 If "Bateman and Shefridge and Wagstaff's conjecture " is
>> true, then all terms of the sequence becomes primes.
>>      %O A000001     3,1
>>     %Y A000001    A000668
>>     %K A000001    nonn
>>     %A A000001    Yasutoshi Kohmoto (zbi74583 at boat.zero.ad.jp)
>>
>>
>>

This seems to be connected to an observation that I made recently,
when I discussed the lengthes of multiplicative cyclic subgroups
base b mod n, i.e., for b=2 and n=7

    0    1   2   3    4    5   6   7  ...  k
    1    2   4   8   16   32  64  128 ...  2^k
    1    2   4   1    2    4          ...  2^k mod 7

L2(7) = 3 since the period-length is 3.

If A(n) = L2(n) then not only the sequence of A(n) is
multiplicative, but also the lengthes L2(p) of primes
seems to be such, that the primefactors of 2^n-1 can
be uniquely determined by the primefactors of (n-1)
and their lengthes. So that mersenne primes do
not only
- need to have prime-exponents in M = 2^p-1, which is widely known ,
but also,
- need, that the cyclic-length (L2(M)=p) is unique.

If we look at p=11 for instance, then 2^11 -1 = 23 * 89;
the cyclic-length of 2^11 is 11; and also the cyclic lengthes
of 23 and 89 are 11. The "uniqueness"-condition, which I assume,
means: they are no other primes with cyclic-length of 11 than
such of the factors of 2^11-1.
Prime are consequently such mersenne-numbers, whose cyclic-length
is unique, like 2^7 -1 = 127 (its cyclic-length is 7); *no* other
prime has this cyclic length, hence 2^7-1 is mersenne-prime.
Consequently, the cyclic-length of primes (base 2) seems determined
basically by the factorization of 2^p-1, and not by other
conditions.

Since the whole sequence A(n) = L2(n) seems multiplicative
(reduced by the least-common-divisor-function, with only the
two exceptions of the known wieferich-primes) the same may
hold for numbers like yours, (2^m_n -2)/3 in a reverse
discussion as a consequence of that, as a consequence of the
above mendtioned determinedness. I can't say at the moment,
if that agrees with the empirical table of values, but I'll
give it a look.


>>     And I knew that it is called "Bateman and Shefridge and Wagstaff's
>>     conjecture ".
>>     Does anyone know the exact description of it?
>>

See http://www.mathdaily.com/lessons/New_Mersenne_conjecture

Regards-

Gottfried Helms