Approximation to the sum of reciprocals of prime numbers squared involving $\phi$, the golden ratio

Asked Feb 16 '21 at 15:30

Active Feb 16 '21 at 15:30

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I was interested in the prime zeta function and its values, so I calculated in Excel the sum of reciprocals of prime numbers squared up to $1$ MM, and tried to relate it to some known irrational numbers ($e$,$\pi$,$\phi$,...). I was surprised when I checked that

$$\frac{\frac{\phi}{2}+1}{4}-\sum_{p<1000000}{1\over p^2}=0.000006896328\dots$$

However, I read here that

$$N=\sum_p{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots$$

And

$$\frac{\frac{\phi}{2}+1}{4}=0.452254\dots$$

¿Is this nice approximation just random? ¿Can someone give a reasonable explanation to this?

asked Feb 16 '21 at 15:30

Juan Moreno

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1

It might be just by chance like here, or here. Or by a series like here. More comments are here. – Dietrich Burde Feb 16 '21 at 15:51
@DietrichBurde I agree with you that most probably it will be by chance, but who know, maybe is related to some interesting series that someone recalls. – Juan Moreno Feb 16 '21 at 17:52
Seems to be a coincidence. But in fact, an astonishing approximation (+1) – Peter Feb 17 '21 at 14:32

Approximation to the sum of reciprocals of prime numbers squared involving $\phi$, the golden ratio

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