Mersenne numbers and the (divisors) property.
Creighton Dement
crowdog at crowdog.de
Tue Apr 26 14:45:21 CEST 2005
> 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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