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%I #42 Mar 21 2023 12:42:54
%S 13,157,913,4513,14647,19201,19291,19813,20191,27778,31828,34825,
%T 37471,39586,40297,50386,53536,53842,54913,62986,64021,70267,76513,
%U 78241,82597,89356,98347,100147,100597,103909,106528,111847,115024,117391,125986,128047
%N Numbers k such that the digits of k^2 are exactly the same (albeit in different order) as the digits of (k+1)^2.
%C All terms are of form 9k+4. - _Zak Seidov_, Jun 04 2010
%C 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
%D Boris A. Kordemsky, The Moscow Puzzles, p. 165 (1972).
%H Paolo P. Lava, <a href="/A072841/b072841.txt">Table of n, a(n) for n = 1..1000</a> (first 519 terms from Zak Seidov)
%e 913 is included because 913^2 = 833569, 914^2 = 835396 and both 833569 and 835396 contain exactly the same set of digits.
%t okQ[n_] := Module[{idn = IntegerDigits[n^2]}, Sort[idn] == Sort[ IntegerDigits[ (n + 1)^2]]]; Select[Range[100000], okQ]
%t 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 *)
%o (PARI) isok(n) = vecsort(digits(n^2)) == vecsort(digits((n+1)^2)); \\ _Michel Marcus_, Sep 30 2016
%K nonn,base
%O 1,1
%A _Harvey P. Dale_, Aug 09 2002
%E Terms from 100147 onward from _N. J. A. Sloane_, May 24 2010
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