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Re: Interesting Mersenne identities and a prime based

• To: mathgroup at smc.vnet.net
• Subject: [mg99923] Re: [mg99887] Interesting Mersenne identities and a prime based
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Tue, 19 May 2009 06:46:46 -0400 (EDT)
• Reply-to: hanlonr at cox.net

And @@ Table[Mod[2^n - 1, 3] == Mod[n, 2], {n, 0, 2000}]

True

And @@ Table[Mod[2^(2^n - 1) - 1, 7] == Mod[n, 2], {n, 0, 29}]

True

Beyond here there be dragons.

Table[{n, Mod[2^(2^(2^n - 1) - 1) - 1, 15] == Mod[n, 2]}, {n, 0, 4}]

{{0, True}, {1, True}, {2, False}, 
   {3, False}, {4, False}}

Table[{n, Mod[2^(2^(2^(2^n - 1) - 1) - 1) - 1, 31] == Mod[n, 2]}, {n, 0, 2}]

{{0, True}, {1, True}, {2, False}}


Bob Hanlon

---- Roger Bagula <rlbagula at sbcglobal.net> wrote: 

=============
My original line of thought  this identity:
Mod[2^n-1,3]

=Mod[n,2]
Are there others?
Table[Mod[2^(2^n - 1) - 1, 7], {n, 1, 10}]
{1, 0, 1, 0, 1, 0, 1, 0, 1, 0}
Mod[2^(2^n - 1) - 1, 7]=Mod[n,2]
What about?
Mod[2^(2^(2^n - 1) - 1) - 1, 15]
Mod[2^(2^(2^(2^n - 1) - 1) - 1) - 1, 31]

That gave me the idea for:
t[n,m]=Mod[Prime[n]^m-2,Prime[n+1]
where
t[n,m]=0
The pairs {n,m} seems to form a connected graph to
n=m=10
at least.

http://www.geocities.com/rlbagulatftn/prime_graph100.gif
The idea of a way of connecting the primes to some graph pattern seems a
new one at least to me.
The graph isn't connected at 20 and appears more random.
At 100 primes which is about what I can get out of my Mathematica
version on my old computer, the graph begins to look like a random
natural network ( the appearance of the primes is sometimes said to be
"normal"). This method at least gives a visualization of that chaos.
It also gives a quantization in terms of spectral graph theory
in which roots can be associated to different levels of primes.
It seems very lucky that I noticed a pattern in the appearance of these
zeros.
Mathematica:
<< DiscreteMath`GraphPlot`;
<< DiscreteMath`ComputationalGeometry`
<< DiscreteMath`Combinatorica`
t[n_, m_] = Mod[Prime[n]^m - 2, Prime[n + 1]]
n0 = 20
g1 = Delete[Union[Flatten[Table[If[t[n, m] == 0, {n, m}, {}], {n, 1,
   n0}, {m, 1, n0}], 1]], 1]
g2 = Table[Reverse[g1[[n]]], {n, 1, Length[g1]}]
Length[%]
g3 = Join[g1, g2]
Length[%]
a = Table[0, {n, n0}, {m, n0}];
Table[If[n == g3[[k, 1]] && m == g3[[k, 2]], a[[n, m]] = 1, {}], {k,
            1, Length[g3]}, {n, n0}, {m, n0}];
a
CharacteristicPolynomial[a, x]
gc60 = FromAdjacencyMatrix[a]
"&#8259;Graph:< 27 20 "Undirected">&#8259;
GraphPlot3D[a]

What I'm wondering is if there is a way
to get a correspondence  of
NSolve[CharacteristicPolynomial[a, x]==0,x]
type roots to the primes involved:
new root<->{ Prime[n],Prime[n+1]}<->{n,m} edge


Roger Bagula