1
$\begingroup$

I'm working on the following sequence of naturals (which is NOT listed in OEIS) $$3,5,11,17,23,29,59,89,119,149,179,209,419,629,839,1049,1259,1469,1679,...$$ whose elements are generated this way $$3=(p_1-1)+p_1\cdot(1)$$ $$5=(p_1-1)+p_1\cdot(p_2-1)$$ $$11=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(1))$$ $$17=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(2))$$ $$23=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(3))$$ $$29=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(p_3-1))$$ $$59=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot((p_3-1)+p_3\cdot(1)))$$ $$...$$ $$209=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot((p_3-1)+p_3\cdot(p_4-1)))$$ $$...$$ where $p_i$ stands for the $i$-th prime. It's quite evident that the sequence is generated by adding the primorials $p_i\#$.

By construction, each element $\,m\,$ of the sequence has the following property: $$m\equiv -1\mod p_i\;$$ for each $\,i\lt k\,$ where $\,k\,$ is the first index such that $\,p_k\# \gt m$.

If we truncate the sequence at the last element generated by using only the first $n$ primes (id est, for example, $209\,$ for $\,n=3$), the whole number of elements is given by $$\sum_{i=1}^{n}(p_{i+1}-1)=(\sum_{i=2}^{n+1}p_i)-n\sim \frac12n^2\log n$$

But what about the number of primes in the sequence (3,5,11,17,23,29,59,89,149,179,419,...)?

Some numeric experiments let me suppose that the number of primes in the truncated sequence above defined could be $\sim 2\,n\log n$.

Is it the correct asymptotical behavior? Could anybody give me the sketch of a proof?

Many thanks.

$\endgroup$
  • 2
    $\begingroup$ It is not clear to me how you construct your sequence ("succession"). Could you explain in words how it's constructed, instead of just listing a bunch of terms? $\endgroup$ – Wojowu Feb 8 at 23:16
  • 1
    $\begingroup$ You start from $3=1+2#$, then add $2#=2$ till you remain under $3#=6$ (that is $5$). Now you continue adding $3#$ unti you remain under $5#=30$. This way you obtain the elements $11,17,23,29$. At this point you have to continue adding $5#$, obtaining $59$ and so on ... $\endgroup$ – Augusto Santi Feb 8 at 23:55
  • $\begingroup$ I checked: the sequence is not present in the OEIS database. $\endgroup$ – Augusto Santi Feb 9 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.