A001292 Concatenations of cyclic permutations of initial positive integers.

1, 12, 21, 123, 231, 312, 1234, 2341, 3412, 4123, 12345, 23451, 34512, 45123, 51234, 123456, 234561, 345612, 456123, 561234, 612345, 1234567, 2345671, 3456712, 4567123, 5671234, 6712345, 7123456

Offset: 1

Author

R. Muller

Keywords

• nonn
• base

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.

Crossrefs

Cf. A093771, A134804, A352991.

Programs

Mathematica

Sort@ Flatten@ Table[ FromDigits[ Join @@ IntegerDigits /@ RotateLeft[Range[n], i - 1]], {n, 11}, {i, n}] (* Giovanni Resta, Mar 21 2017 *)

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