

A322522


a(n) is the minimal absolute difference between n and each of the powers of the previous terms; a(1) = 1.


1



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, 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, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1
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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 (i1)^2 and i^2. We then have a((i1)^2 + i) = i = ceiling(sqrt((i1)^2 + i)). Therefore there are infinitely many numbers n such that a(n) = ceiling(sqrt(n)).
a(n)  a(n1) <= 1.


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

Cf. A301573 (distance from n to the nearest nontrivial power).
Sequence in context: A293388 A268833 A293386 * A168121 A158948 A140224
Adjacent sequences: A322519 A322520 A322521 * A322523 A322524 A322525


KEYWORD

nonn,easy,look,nice,hear


AUTHOR

Gabin Kolly, Aug 28 2019


STATUS

approved



