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A308493
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Numbers k such that k in base 10 contains the same digits as k in some other base.
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2
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 20, 21, 23, 31, 41, 42, 43, 46, 51, 53, 61, 62, 63, 71, 73, 81, 82, 83, 84, 86, 91, 93, 100, 101, 102, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 121, 122, 123, 131, 133, 141, 144, 151, 155, 158, 161, 166, 171, 177, 181
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OFFSET
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1,3
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COMMENTS
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This sequence is infinite because 2*10^k is a term for any k >= 0.
Also 10^k is a term when k >= 0 and so too 10^k*(10^m - 1)/9 for any k > 0 and m >= 0. - Bruno Berselli, Aug 26 2019
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LINKS
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EXAMPLE
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k = 113 is in the sequence because the set of digits of k {1, 3} equals the set of digits of (k in base 110) = 13.
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PROG
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(PARI) isok(k) = {my(j=Set(digits(k))); for(b=2, k+1, if((b!=10) && (Set(digits(k, b)) == j), return(1))); return(0); } \\ Michel Marcus, Aug 05 2019
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CROSSREFS
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Cf. A034294, A037415, A037422, A037428, A037433, A037437, A037440, A037442, A037443, A249899, A307498.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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