OFFSET
1,3
COMMENTS
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)).
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)).
a(n) - a(n-1) <= 1.
For the first 10000 terms, indices for a(n) = 1 correspond to 85 of 90 values of A227802(m). - Bill McEachen, Feb 26 2024
LINKS
Gabin Kolly, Table of n, a(n) for n = 1..10000
FORMULA
Let b(n) be the first time that n appears in the sequence; then b(n) ~ n^2.
EXAMPLE
For n = 4, we have a(4) = 0, because a(3) = 2, and 2^2 - 4 = 0.
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.
MATHEMATICA
comparePowers[n_, m_] :=
If[n <= 1, m - n, a = n; While[a < m, a *= n];
Min[m - a/n, a - m]]; list = {1}; cleanList = {1}; Do[
list = Append[list,
Min[comparePowers[#, Length[list] + 1] & /@ cleanList]];
If[Last[list] > Last[cleanList],
cleanList = Append[cleanList, Last[list]]], 9999]; Print[list]
CROSSREFS
AUTHOR
Gabin Kolly, Aug 28 2019
STATUS
approved