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%I #48 Mar 23 2024 21:05:30
%S 1,1,2,0,1,2,1,0,1,2,3,3,3,2,1,0,1,2,3,4,5,3,2,1,0,1,0,1,2,2,1,0,1,2,
%T 3,4,5,6,3,4,5,6,7,5,4,3,2,1,0,1,2,3,4,5,6,7,7,6,5,4,3,2,1,0,1,2,3,4,
%U 5,6,7,8,8,7,6,5,4,3,2,1,0,1
%N a(n) is the minimal absolute difference between n and each of the powers of the previous terms; a(1) = 1.
%C a(n) <= ceiling(sqrt(n)); if n is between k^2 and (k+1)^2, we have min(n - k^2, (k+1)^2 - n) <= k < ceiling(sqrt(n)).
%C Using the fact that the density of the nontrivial powers over the integers is 0, and that the density of cubes and higher powers among the nontrivial powers is 0, we can show that there are an infinite number of integers i such that i is not a nontrivial power, and there is no cube or higher power between (i-1)^2 and i^2. We then have a((i-1)^2 + i) = i = ceiling(sqrt((i-1)^2 + i)). Therefore there are infinitely many numbers n such that a(n) = ceiling(sqrt(n)).
%C a(n) - a(n-1) <= 1.
%C For the first 10000 terms, indices for a(n) = 1 correspond to 85 of 90 values of A227802(m). - _Bill McEachen_, Feb 26 2024
%H Gabin Kolly, <a href="/A322522/b322522.txt">Table of n, a(n) for n = 1..10000</a>
%F Let b(n) be the first time that n appears in the sequence; then b(n) ~ n^2.
%e For n = 4, we have a(4) = 0, because a(3) = 2, and 2^2 - 4 = 0.
%e For n = 6, we have a(6) = 0, because there are only 0, 1 and 2 in the first 5 terms, and therefore the closest power is 2^2 = 4 or 2^3 = 8, with an absolute difference of 2.
%t comparePowers[n_, m_] :=
%t If[n <= 1, m - n, a = n; While[a < m, a *= n];
%t Min[m - a/n, a - m]]; list = {1}; cleanList = {1}; Do[
%t list = Append[list,
%t Min[comparePowers[#, Length[list] + 1] & /@ cleanList]];
%t If[Last[list] > Last[cleanList],
%t cleanList = Append[cleanList, Last[list]]], 9999]; Print[list]
%Y Cf. A301573 (distance from n to the nearest nontrivial power).
%K nonn,easy,look,nice,hear
%O 1,3
%A _Gabin Kolly_, Aug 28 2019
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