Problems & Puzzles: Conjectures

Conjecture 108. On the counting function of semiprimes 

On Dec. 30, Alain Rocchelli sent the following conjecture:

In OEIS A001358 (Numbers of the form p*q where p and q are primes), it is mentioned:

"However, the asymptotic formula shows that the linearity is an illusion and in fact a(n)/n ~ log(n)/log(log(n)) goes to infinity". Here, log(e)=1.

Based on expansive computations, integrating the data from A066265 and the formula of Dragos Crisan and Radek Erban, On the counting function of semiprimes, it seems that this formula can be improved by modifying it:

a(n)/n ~ [log(n)/log(log(n))]*{1-(0.08/sqrt[log(log(n))])} which remains equivalent for n --> infinity.

Q1. Do you agree with this new formula?

Q2. Can you get a better formula, especially for the value 0.08?


On Jan 29, 2025, Alain Rochelli wrote:

After much thought, it seems that this conjecture is not entirely relevant but it helps to numerically verify Landau's theorem.

To allow very heavy and intensive computations with GP PARI (or other applications), it is judicious to be interested in the local form of this conjecture:

By derivation of the simplified formula of Dragos Crisan and Radek Erban we obtain:

{DN/DX} (X*log(log(X)/log(X))+0.265*X/log(X) ~ log(log(X))/log(X)+0.265/log(X) = [log(log(X)+0.265] / log(X)

with N as the number of semiprimes less than X.

It follows that: DX/DN ~ log(X) / [log(log(X))+0.265]

Similarly for X ~ C*N*log(N)/log(log(N)) with C considered as constant:

DX/DN = C * [ (log(N)+1) / log(log(N)) - 1/(log(log(N)))^2 ] ~ C * log(N) / log(log(N)) for N (and X) --> infinity

By computation, we get with C = {1-[A/sqrt(log(log(N)))]} :


X=10^100 ; DX/DN=40.36647 ; C=0.97302 ; A=0.06675

X=10^200 ; DX/DN=71.98559 ; C=0.96609 ; A=0.08391

X=10^400 ; DX/DN=129.8970 ; C=0.96699 ; A=0.08621

X=10^800 ; DX/DN=236.6588 ; C=0.96845 ; A=0.08650

X=10^1600 ; DX/DN=434.6145 ; C=0.97014 ; A=0.08555

X=10^3200 ; DX/DN=803.5247 ; C=0.97188 ; A=0.08390

X=10^6400 ; DX/DN=1494.111 ; C=0.97356 ; A=0.08319

X=10^12800 ; DX/DN=2792.008 ; C=0.97513 ; A=0.08211

X=10^25600 ; DX/DN=5239.949 ; C=0.97657 ; A=0.08083

We can see that C converges to 1, but A seems to decrease very very slowly to 0.

 

***

On Feb 9, 2025 Alain added:

I complete my previous contribution by proposing a better formula for a(n)/n=X/N, namely:

X/N ~ [log(N)/log(log(N))] * {1-[M/log(log(N))]} where M is the Mertens constant (0.261497..)

The explanation is obtained by proceeding as before but this time starting from the slightly different formula:

N ~ (X*log(log(X))/log(X)) + 0.261497*X/log(X) ; cf. formula 17 of the publication by D. Crisan and R. Erban

We then obtain:

DN/DX ~ [log(log(X))+0.261497] / log(X) with N as the number of semiprimes less than X.

It follows that: DX/DN ~ log(X) / [log(log(X))+0.261497].

Now, for X (and N) --> infinity, we can verify that: log(X)/log(log(X)) ~ log(N)/log(log(N)).

Indeed, by direct application of Landau's theorem and by its derived form, we obtain:

X ~ N*log(N)/log(log(N)) involving DX/DN ~ log(N)/log(log(N)) and by combining the two:

X/N ~ DX/DN involving DX/X ~ DN/N

Recognizing the logarithmic derivation, we get log(X) ~ log(N) and log(log(X)) ~ log(log(N))

Then, reconsidering, as before, DX/DN ~ C * log(N)/log(log(N)) for N (and X) --> infinity.

It follows that:

DX/DN ~ log(X)/[log(log(X))+0.261497] ~ log(N)/[log(log(N))+0.261497] ~ C*log(N)/log(log(N))

By setting C = 1-[M/log(log(N))], we obtain the value 0.261497 for M as N tends to infinity.

 

 

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