valuation(2, prime(n)-1)

[benoit]

dear Pieter Moree,

Many thanks for this instructive answer and providing references.

>> Letting S(n)=sum(i=2, n, valuation(2, prime(i)-1) )
>>
>> I suppose limit n-->infty S(n)/n = 2
> This is true and is not too difficult to prove:
>
> The primes (*) p=1(mod 2^m) that are not =1(mod 2^{m+1})
>
> have
> density 2^{-m} and contribute m/2^m to the density.
> Summing this for m is 1 to infty then yields the result.
> (This is the idea, in the proof one ought be a bit more careful.)


More genrally, letting S(n,p)= sum(i=2, n, valuation(p, prime(i)-1) )

If I'm not totally wrong, similar arguments should give :

limit S(n,p)/n = sum(m>=1,m/p^m)=p/(p-1)^2

limit S(n,2)/n=2
limit S(n,3)/n=3/4
limit S(n,5)/n=5/16
limit S(n,7)/n=7/36

>> I asked myself if (2n-S(n))/sqrt(n) is bounded with S(n)<2n
> In my opinion this is rather too optimistic. If one assumes
> the Generalized Riemann Hypothesis then one can show there
> is a constant c such that if one replaces aqrt(n) by
> sqrt(n)(log n)^c, then the quotient is bounded.
>> and if limit n-->infty (2n-S(n))/sqrt(n) exist ?
> I am convinced the answer is: NO. S(n) is quite irregular.
> The analog to think of is: (pi(x)-Li(x))/sqrt{x}.
> It can be proved that this does not have a limit.

It's clear from Littlewood. I wonder what is the first n for which 
S(n)>2n ?

Following your analogy : why not conjecture c=1 and :

S(n)=2n+O(n^(1/2)*log(x))

making the full parallel with :

pi(n)=Li(n)+O(n^(1/2)*log(n)) if RH is true.

Or for any p :

S(n,p)=p/(p-1)^2*n+O(n^(1/2)*log(x))


> Presumably variations of these techniques can be used to deal with
> the latter two questions conclusively.

You are certainly right. For exemple, p/(p-1)^2*n-S(n,p) has changes of 
sign for p>2 :

for p=3 first for n=518
for p=5 first for n=54


>> I don't believe that's specific to primes but thanks for confirming 
>> or not this fact.
> Oh yes it is ! Instead of the primes take the odd integers <=x. A 
> different
> picture emerges there ! The fluctations will be quite small now 
> (indeed bounded by a constant times (log x)^2...)
>
> If instead of primes one takes primes =a(mod d) with a and d coprime 
> and d odd, then the same
> features should arise, however.

I tried sigma(n) and it is also very different.

Thanks again,
Benoit Cloitre
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