OFFSET
1,2
COMMENTS
Entries are sorted numerically, so after a(45) = 912345678 we have a(46) = 10123456789 instead of a(46) = 12345678910. - Giovanni Resta, Mar 21 2017
From Marco Ripà, Apr 21 2022: (Start)
In 1996, Kenichiro Kashihara conjectured that there is no prime power of an integer (A093771) belonging to this sequence (disregarding the trivial case 1); a direct search from 12 to a(100128) has confirmed the conjecture up to 10^1035. There are no perfect powers among terms t which are permutations of 123_...(m - 1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, any perfect power in this sequence is necessarily congruent modulo 9 to 0 or 1.
(End)
LINKS
John Cerkan, Table of n, a(n) for n = 1..10000
Marco Ripà, On some open problems concerning perfect powers, ResearchGate (2022).
Florentin Smarandache, Only Problems, Not Solutions!, Unsolved Problem #16, p. 18.
MATHEMATICA
Sort@ Flatten@ Table[ FromDigits[ Join @@ IntegerDigits /@ RotateLeft[Range[n], i - 1]], {n, 11}, {i, n}] (* Giovanni Resta, Mar 21 2017 *)
PROG
(Python)
from itertools import count, islice
def A001292gen():
s = []
for i in count(1):
s.append(str(i))
yield from sorted(int("".join(s[j:]+s[:j])) for j in range(i))
print(list(islice(A001292gen(), 46))) # Michael S. Branicky, Jul 01 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
R. Muller
STATUS
approved