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A307498
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Numbers k such that the digits of k in base 10 are a permutation of those of k in some other base.
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3
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 21, 23, 31, 41, 42, 43, 46, 51, 53, 61, 62, 63, 71, 73, 81, 82, 83, 84, 86, 91, 93, 158, 191, 196, 227, 261, 265, 283, 316, 370, 371, 441, 445, 511, 518, 551, 774, 782, 825, 834, 882, 910, 911, 912, 913, 914, 915, 916, 917, 918
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OFFSET
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1,3
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COMMENTS
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If the digits of k in base 10 is a permutation of m = (k in base b), 10^j < k < 10^(j+1), then 10^(j/(j+1)) < b < 10^((j+1)/j).
If k > 10, other base can only be 4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 25, 26, 28, 37, 46, 55, 64, 73, 82.
The digits of k in base 10 is a permutation of k in base 82 iff k = 91.
The largest term is less than 10^25. See proof in A034294.
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LINKS
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EXAMPLE
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13 in base 4 is 31, 227 in base 9 is 272.
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PROG
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(PARI) isok(k) = {my(v = vecsort(digits(k))); k < 10 || sum(j = 3, 82, vecsort(digits(k, j)) == v) > 1; }
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CROSSREFS
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KEYWORD
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nonn,base,fini
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AUTHOR
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STATUS
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approved
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