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A001292
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Concatenations of cyclic permutations of initial positive integers.
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6
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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
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OFFSET
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1,2
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COMMENTS
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Entries are sorted numerically, so after a(45) = 912345678 we have a(46) = 10123456789 instead of a(46) = 12345678910. - Giovanni Resta, Mar 21 2017
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)
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LINKS
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MATHEMATICA
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Sort@ Flatten@ Table[ FromDigits[ Join @@ IntegerDigits /@ RotateLeft[Range[n], i - 1]], {n, 11}, {i, n}] (* Giovanni Resta, Mar 21 2017 *)
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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R. Muller
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STATUS
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approved
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