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[Solved] Summation of primes and nonprimes with same result

Why the summation of the given formula gives the same results, if in one I use primes and in the other I don't?

sq=Table[j,{j,1000}]
n=Select[sq,PrimeQ,(100)]
sq2=Table[k,{k,100}]
n3=sq2*-1
r=Table[k1,{k1,100}]
f=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n]))
bb=Im[f]
s1cc=(((1)+bb*r*Sqrt[-1])+((0)+bb*r*Sqrt[-1]))/2
zz=-n3
zx=n
x1c1c=\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(zx = 1\), \(100\)]\((1/zx*zx^s1cc)\)\)`
`x1c1c2=\!\(

x1c1c2=\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(zz = 1\), \(100\)]\((1/zz*zz^s1cc)\)\)
2 Replies
Anonymous User
Anonymous User
Posted 3 years ago

I think the confusion is due to your assigning zz and zx in the Sum as increments which eliminates the previous assignment and makes them equal.

$\sum _{j=1}^2 \frac{\text{zz}[[j]]^{\text{s1cc}}}{\text{zz}[[j]]}$

Having tried this upon both they are no longer equal and appear like you had suspected.

POSTED BY: Anonymous User

Thank you so much John! You confirmed me what I thought it was.

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