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%I #20 Aug 12 2022 19:49:14
%S 32,18,8,350,480,21,104,125874,40,4147,5712,65,15680,585,96,833,7776,
%T 133,61600,46851,176,63595,2232,225,106288,122931,280,3857,91948380,
%U 341,237184,83853,408,543900,7236,481,8474,431516631,560,622339,683760,645
%N a(n) is least k such that k and 2k are anagrams in base n (written in base 10).
%C a(3n-1) = A000567(n). - _Peter Kagey_, Dec 16 2016
%C a(n) is divisible by n-1 for all n due to the anagram property: n and 2n are congruent modulo n-1. - _Peter Kagey_, Dec 23 2016
%H Mathematics Stack Exchange, <a href="http://math.stackexchange.com/questions/2061958">Lower Bound for Anagram Sequence</a>.
%t Table[k = 1; While[! MatchQ @@ Map[Sort@ IntegerDigits[#, n] &, k {1, 2}], k++]; k, {n, 3, 30}] (* _Michael De Vlieger_, Dec 17 2016 *)
%o (PARI) a(n) = {my(k=1); while (vecsort(digits(k,n)) != vecsort(digits(2*k,n)), k++); k;} \\ _Michel Marcus_, Dec 17 2016
%Y Cf. A023095, A023096, A023097, A023098, A023099, A023100, A023101, A023102.
%K nonn,base
%O 3,1
%A _David W. Wilson_
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