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A072841
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Numbers k such that the digits of k^2 are exactly the same (albeit in different order) as the digits of (k+1)^2.
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6
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13, 157, 913, 4513, 14647, 19201, 19291, 19813, 20191, 27778, 31828, 34825, 37471, 39586, 40297, 50386, 53536, 53842, 54913, 62986, 64021, 70267, 76513, 78241, 82597, 89356, 98347, 100147, 100597, 103909, 106528, 111847, 115024, 117391, 125986, 128047
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OFFSET
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1,1
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COMMENTS
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All terms are of form 9k+4. - Zak Seidov, Jun 04 2010
All numbers of the form 5500*10^k - 87, k >= 1 are terms, i.e., 54 followed by k 9's followed by a 13: 54913, 549913, 5499913, etc. - Enrico Munini, Feb 21 2023
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REFERENCES
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Boris A. Kordemsky, The Moscow Puzzles, p. 165 (1972).
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LINKS
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EXAMPLE
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913 is included because 913^2 = 833569, 914^2 = 835396 and both 833569 and 835396 contain exactly the same set of digits.
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MATHEMATICA
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okQ[n_] := Module[{idn = IntegerDigits[n^2]}, Sort[idn] == Sort[ IntegerDigits[ (n + 1)^2]]]; Select[Range[100000], okQ]
SequencePosition[Table[FromDigits[Sort[IntegerDigits[n^2]]], {n, 130000}], {x_, x_}][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 09 2020 *)
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PROG
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(PARI) isok(n) = vecsort(digits(n^2)) == vecsort(digits((n+1)^2)); \\ Michel Marcus, Sep 30 2016
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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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EXTENSIONS
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STATUS
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approved
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