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%I #21 Aug 08 2014 14:38:06
%S 0,0,3,3,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N a(n)= 1 (respectively, a(n)= 3) if up to 2^n the number of A245622-terms is more (respectively, less) than the number of A245623-terms; or a(n)=0 if these numbers are equal.
%C Since 10^2,10^3,... appear earlier than 3*10^2,3*10^3,..., then it is natural to conjecture, that, beginning with n=4, a(n)=0 or 1.
%o (PARI) conv(n) = if (n==0, 0, if (n>0, 3, 1));
%o nbdv(v, d) = my(m=0); for(i=1, #v, if(v[i]==d, m++)); m;
%o a(n) = {nb3 = 0; nb1 = 0; forprime (j=1, 2^n, v = digits(j); d3 = nbdv(v, 3); d1 = nbdv(v, 1); if (d3 > d1, nb3++); if (d1 > d3, nb1++);); conv(nb3 - nb1);} \\ _Michel Marcus_, Aug 04 2014
%Y Cf. A245622, A245623.
%K nonn,base,more
%O 0,3
%A _Vladimir Shevelev_, Jul 27 2014
%E More terms from _Peter J. C. Moses_, Jul 27 2014
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