%I #19 Dec 11 2023 08:38:58
%S 843,1043,1083,1091,2043,2083,2091,2283,2291,2331,3343,3543,3583,3591,
%T 3843,4093,4143,4193,4203,4213,4215,4217,4219,4223,4243,4343,4593,
%U 4793,4833,4841,5043,5083,5091,5143,5193,5203,5213
%N Numbers whose base-5 representation contains exactly two 1's and three 3's.
%H Robert Israel, <a href="/A045243/b045243.txt">Table of n, a(n) for n = 1..10000</a>
%p A0[1]:= {2,4}:
%p A1[1]:= {1}:
%p A3[1]:= {3}:
%p A11[1]:= {}:
%p A13[1]:= {}:
%p A33[1]:= {}:
%p A113[1]:= {}:
%p A133[1]:= {}:
%p A333[1]:= {}:
%p A1133[1]:= {}:
%p A1333[1]:= {}:
%p A11333[1]:= {}:
%p count:= 0:
%p for n from 2 while count < 1000 do
%p A0[n]:= map(t -> (5*t, 5*t+2, 5*t+4), A0[n-1]);
%p A1[n]:= map(t -> (5*t,5*t+2,5*t+4), A1[n-1]) union map(t -> 5*t+1, A0[n-1]);
%p A3[n]:= map(t -> (5*t,5*t+2,5*t+4), A3[n-1]) union map(t -> 5*t+3, A0[n-1]);
%p A13[n]:= map(t -> (5*t,5*t+2,5*t+4), A13[n-1]) union map(t -> 5*t+1, A3[n-1]) union map(t -> 5*t+3, A1[n-1]);
%p A33[n]:= map(t -> (5*t,5*t+2,5*t+4), A33[n-1]) union map(t -> 5*t+3, A3[n-1]);
%p A113[n]:= map(t -> (5*t,5*t+2,5*t+4), A113[n-1]) union map(t -> 5*t+1, A13[n-1]) union map(t -> 5*t+3, A11[n-1]);
%p A133[n]:= map(t -> (5*t,5*t+2,5*t+4), A133[n-1]) union map(t -> 5*t+1, A33[n-1]) union map(t -> 5*t+3, A13[n-1]);
%p A333[n]:= map(t -> (5*t,5*t+2,5*t+4), A333[n-1]) union map(t -> 5*t+3, A33[n-1]);
%p A1133[n]:= map(t -> (5*t,5*t+2,5*t+4), A1133[n-1]) union map(t -> 5*t+1, A133[n-1]) union map(t -> 5*t+3, A113[n-1]);
%p A1333[n]:= map(t -> (5*t,5*t+2,5*t+4), A1333[n-1]) union map(t -> 5*t+1, A333[n-1]) union map(t -> 5*t+3, A133[n-1]);
%p A11333[n]:= map(t -> (5*t,5*t+2,5*t+4), A11333[n-1]) union map(t -> 5*t+1, A1333[n-1]) union map(t -> 5*t+3, A1133[n-1]);
%p count:= count + nops(A11333[n]);
%p od:
%p sort([seq(op(A11333[i]),i=1..n-1)]); # _Robert Israel_, Dec 10 2023
%t Select[Range[6000],DigitCount[#,5,1]==2&&DigitCount[#,5,3]==3&] (* _Harvey P. Dale_, Feb 25 2013 *)
%Y Cf. A007091.
%K nonn,base
%O 1,1
%A _Clark Kimberling_