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k and 2k are anagrams in base 5 (written in base 5).
1

%I #10 Aug 08 2018 04:29:36

%S 13,130,143,1300,1313,1430,1443,13000,13013,13130,13143,14300,14313,

%T 14430,14443,102342,103242,120234,120324,130000,130013,130130,130143,

%U 131300,131313,131430,131443,143000,143013,143130,143143,144300,144313

%N k and 2k are anagrams in base 5 (written in base 5).

%C From _Robert Israel_, Aug 08 2018: (Start)

%C The concatenation of two terms is a term.

%C If a*10^m + b is a term, where b < (2/9)*10^m, then a*10^k+b is a term for all k > m. (End)

%H Robert Israel, <a href="/A023061/b023061.txt">Table of n, a(n) for n = 1..2000</a>

%p f:= proc(n) local L, M;

%p L:= convert(n, base, 5);

%p M:= convert(2*n, base, 5);

%p if sort(L) = sort(M) then add(L[i]*10^(i-1), i=1..nops(L)) else NULL fi

%p end proc:

%p map(f, [$1..10000]); # _Robert Israel_, Aug 08 2018

%K nonn,base

%O 1,1

%A _David W. Wilson_