Mersenne numbers and the (divisors) property.

Gerald McGarvey Gerald.McGarvey at comcast.net
Wed Apr 27 15:52:50 CEST 2005


Something to note about A007733:
For 2 <= n <= 84, A000010(n) / A007733(n) results in the following sequence:
1 1 2 1 1 2 4 1 1 1 2 1 2 2 8 2 1 1 2 2 1 2 4 1 1 1 4 1 2 6 16 2 2 2 2 1 1 
2 4 2 2 3 2 2 2 2 8 2 1 4 2 1 1 2 8 2 1 1 4 1 6 6 32 4 2 1 4 2 2 2 4 8 1 2 
2 2 2 2 8 1 2 1 4
and it seems reasonable to conjecture that A007733 always divides A000010 
(Euler totient function phi(n)).
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000010
-- Gerald

At 09:32 AM 4/27/2005, Gerald McGarvey wrote:

>A related entry the Mersenne numbers is A007733 (Period of binary 
>representation of 1/n.)
>http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A007733
>"
>Also sequence of period lengths for n's when you do
>primality testing and calculate
>"2^k mod n" from k=0 to k=n - Gottfried Helms
>(helms(AT)uni-kassel.de), Oct 05 2000"
>
>Sincerely,
>Gerald
>
>At 08:45 AM 4/26/2005, Creighton Dement wrote:
>> > Dear Seqfans,
>> >
>> > I would like to mention two specific examples concerning the property,
>> > for m > n : if s | a(n) and s | a(m) then s | a(2*m - n)
>> >
>> > A proof is given, below, in an attempt to show that the first example
>> > has the property, the second doesn't (although, in a way, it "almost"
>> > seems to...).
>> >
>> > Ex. 1.
>> > a(n) = a(n-1) + 2a(n-2) + 2, a(0) = 0, a(1) = 1.
>> > Ex. 2.
>> > a(n) = a(n-1) + 2a(n-2) + 3, a(0) = 0, a(1) = 1.
>> >
>> > Ex. 1., Mersenne numbers
>> > http://www.research.att.com/projects/OEIS?Anum=A000225
>> > Sequence:
>> > [0,1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,
>> > 32767,65535,131071,262143,524287,1048575,2097151,4194303,
>> > 8388607,16777215,33554431,67108863,134217727,268435455,
>> > 536870911,1073741823,2147483647,4294967295]
>> >
>> > Factorized:
>> > [0, 1, (3), (7), (3)*(5), (31), (3)^2*(7), (127), (3)*(5)*(17),
>> > (7)*(73), (3)*(11)*(31), (23)*(89), (3)^2*(5)*(7)*(13),
>> > (8191),(3)*(43)*(127), (7)*(31)*(151), (3)*(5)*(17)*(257), (131071),
>> > (3)^3*(7)*(19)*(73), (524287), (3)*(5)^2*(11)*(31)*(41),
>> > (7)^2*(127)*(337), (3)*(23)*(89)*(683), (47)*(178481),
>> > (3)^2*(5)*(7)*(13)*(17)*(241), (31)*(601)*(1801), (3)*(8191)*(2731),
>> > (7)*(73)*(262657), (3)*(5)*(29)*(43)*(113)*(127), (233)*(1103)*(2089)]
>> >
>> > Notice, for example, that the number 17 seems to reappear every 8
>> > terms.
>> > The number 3 every 2 terms, etc.. (This apparently defines a function
>> > f such that f(17) = 3, f(3) = 2, etc. which is beside the point for
>> > now.).
>> > Also note that the fact that the number 17 actually appears for the
>> > first time after 8 terms, etc., (i.e. a(8) = 17) is not covered by the
>> > property- that seems to be an additional "symmetry" for this
>> > particular sequence (which would be in need of an additional proof!).
>>
>>I assume that the following is the last sentence reformulated as a
>>conjecture:
>>for m > n: if s | 2^n - 1 and s | 2^m - 1 then
>>(m - n) | A002326( (s-1)/2 )
>>
>>http://www.research.att.com/projects/OEIS?Anum=A002326
>>( least m such that 2n+1 divides 2^m-1 )
>>
>>Not sure if that conjecture is somehow enlightening, obvious, or would
>>only serve to confuse - thus I will hold off on posting it as a comment
>>for now.
>>
>>Note: Correction, the above should read f(17) = 8. (I also accidently
>>gave the proof for 2^n+1 instead of 2^n - 1; essentially, both proofs
>>are the same)
>>
>>Sincerely,
>>Creighton
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