A301573 -id:A301573 - cp's OEIS frontend

A366928 a(n) is the smallest nonnegative k such that A301573(k) = n.

1, 0, 6, 12, 20, 41, 42, 56, 72, 90, 110, 155, 156, 182, 270, 271, 272, 306, 379, 380, 420, 462, 551, 552, 600, 650, 702, 756, 812, 870, 930, 1055, 1056, 1122, 1190, 1260, 1405, 1406, 1482, 1560, 1640, 1805, 1806, 1892, 1980, 2254, 2255, 2256, 2352, 2450, 2550, 2652, 2861, 2862, 2970

Offset: 0

Dmitry Kamenetsky, Oct 28 2023

nonn

If negative values were allowed then one could argue a(n) = 1-n from the name of A301573. - David A. Corneth, Nov 13 2023

			a(3) = 12 as 12 is the smallest positive integer that is 3 away from the closest perfect power (namely 9 = 3^2). - _David A. Corneth_, Nov 12 2023

David A. Corneth, Table of n, a(n) for n = 0..10000
David A. Corneth, PARI program

Cf. A001597, A301573.

ispp(n) = {ispower(n) || n==1}; \\ A001597
f(n) = my(k=0); while(!ispp(n+k) && !ispp(n-k), k++); k; \\ A301573
a(n) = my(k=0); while (f(k) != n, k++); k; \\ Michel Marcus, Oct 29 2023

PARI
```
\\ See PARI link
```

from itertools import count
from sympy import perfect_power
def A366928(n): return next(m for m in count(0) if next(k for k in count(0) if perfect_power(m+k) or perfect_power(m-k) or m-k==1 or m+k==1) == n) # Chai Wah Wu, Nov 12 2023

a(n) > n^2 for n > 1. - David A. Corneth, Nov 13 2023

More terms from Michel Marcus, Oct 29 2023

A322522 a(n) is the minimal absolute difference between n and each of the powers of the previous terms; a(1) = 1.

Original entry on oeis.org

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

Offset: 1

Gabin Kolly, Aug 28 2019

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

			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.

Gabin Kolly, Table of n, a(n) for n = 1..10000

Cf. A301573 (distance from n to the nearest nontrivial power).

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]

Let b(n) be the first time that n appears in the sequence; then b(n) ~ n^2.

cp's OEIS Frontend

A366928 a(n) is the smallest nonnegative k such that A301573(k) = n.

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