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A136646 Primes that give find root imaginary results for: z1 = 1/2 - 4*I*Pi*x; z2 = 1 - 8*I*Pi*x; 1 + 2 x^z1 + x^z2 == 0; with Im[x]<-0.05. The resulting root share value of the Zeta[root] being near one. 0

%I #5 Mar 01 2013 06:43:12

%S 89,619,877,1193,1319,1481,1747,2437,2447,2521,2647,3163,3371,3449,

%T 3469,5711,6007,6089,6659,7001,7121

%N Primes that give find root imaginary results for: z1 = 1/2 - 4*I*Pi*x; z2 = 1 - 8*I*Pi*x; 1 + 2 x^z1 + x^z2 == 0; with Im[x]<-0.05. The resulting root share value of the Zeta[root] being near one.

%C This result is my second try at this type of experiment. I had this idea of looking at the product elements in the zeta product as limiting near primes.

%C pe[n]=1/(1-Prime[n]^(-z)) ; Limit[Pe[n],x->Prime[n]+Delta1+I*Delta2]=0, where z=-1/2+i*4*Pi*Prime[n].

%C I solved it down to an equation in n x and then, I looked a near specific primes. Two types show up: Normal Riemannian -1/2 primes on a curve and second type that are attracted to one instead.

%C There seems to be a spectral effect in the Delta2 values. The curve ZDelta2 go to a lower limit and the second types are all well below that limiting curve.

%F x start at Prime[n]: when the equation z1 = 1/2 - 4*I*Pi*x; z2 = 1 - 8*I*Pi*x; 1 + 2 x^z1 + x^z2 == 0; has a root with Im[x]<-0.05. the starting prime is reported out.

%t z1 = 1/2 - 4*I*Pi*x; z2 = 1 - 8*I*Pi*x; a1 = Flatten[Table[If[(Im[x] /. FindRoot[1 + 2 x^z1 + x^z2 == 0, {x, Prime[n]}]) < -0.05, Prime[n], {}], {n, 1, 1000}]]

%K nonn,uned

%O 1,1

%A _Roger L. Bagula_, Apr 14 2008

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