1
$\begingroup$

Pair of sequences $\ v_n\ $ and $\ U_n\ $ of integers start as in the following table:

[\begin{array}{rrrrrrrrrr} n= & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \ldots \\ v_n= & 0 & 2 & 5 & 10 & 17 & 37 & 50 & 82 & \ldots \\ U_n= & 0 & 2 & 3 & 6 & 8 & 12 & 14 & 18 & \ldots \end{array}]

These two sequences are defined as follows:

  • $\ v_0=U_0=0;$
  • $\ v_n\in\mathbb N\ $ is the smallest natural number such that none of the consecutive $\,\ U_{n-1}\!+\!1\,\ $ integers $\ v_n\ \ldots\ v_n\!+\!U_{n-1}\ $ is powerful;
  • $\ U_n\ $ is the smallest natural number such $\ v_n+U_n\ $ is powerful.

Thus, we are looking at the ever longer maximal sequences of consecutive non-powerful sequences. One would like to know the behavior of these sequences:

Question:   what are reasonable (as exact as possible, and easily computable) lower and upper bounds for terms $\ v_n\ $ and $\ U_n,\ $ and their asymptotic behavior?

Knowing roughly the number of powerful initegers $\ POW(x)\ $ that do not exceed $\ x\ $ (for every positive $\ x\in\mathbb R),\ $ we may deduce the average behavior of these sequences of non-powerful integers; the still harder challenge would be deducing the more delicate but consistent deviations from the regular statistical behavior.

$\endgroup$
2
  • $\begingroup$ Sequence/segment $\ 2,3,6,8,12,14,18\ $ appears here: oeis.org/…. This may (or may not) be accidental. I'll write a simple Perl program, it will expand the initial segments of $\ v_n\ $ and $\ U_n$. $\endgroup$
    – Wlod AA
    Oct 31, 2022 at 3:24
  • 1
    $\begingroup$ OEIS A001694 links to Paul T. Bateman and Emil Grosswald, On a theorem of Erdős and Szekeres, Illinois J. Math. 2:1 (1958), pp. 88-98, and notes that it gives $\textrm{POW}(x) = \frac{\zeta(\tfrac 32)}{\zeta(3)} x^{\tfrac12} + \frac{\zeta(\tfrac 23)}{\zeta(2)} x^{\tfrac13} + o(x^{\tfrac16})$ $\endgroup$ Oct 31, 2022 at 12:05

1 Answer 1

4
$\begingroup$

Infinitely often (and with positive density, as proved by Shiu) there are no powerful numbers between $n^2+1$ and $(n+1)^2-1$. Hence the maximal gap below $x$ is infinitely often of size $\sim2\sqrt x$. (This is more than can be deduced from the Erdős & Szekeres result, or the Bateman & Grosswald result, since $\zeta(3/2)/\zeta(3)>2.$)

$U_n$ is largest if $v_n$ is a square and the above interval is empty. So $$ U_n \le (\sqrt{v_n}+1)^2 - v_n = 2\sqrt{v_n}+1 $$ with equality holding in that case. Deducing a lower bound is probably tantamount to finding gaps in A336175, but I would be surprised if it was not $\sim2\sqrt{v_n}$ as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.