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%I #6 Mar 30 2012 17:28:13
%S 679922,679923,679924,679925,679926,679927,679928,679929,1048576,
%T 16777216,17457138,17457139,17457140,17457141,17457142,17457143,
%U 17457144,17457145,17825792
%N n is equal to the number of 2's in all numbers <= n written in base 8.
%o (Perl) ($s,$t,$u)=(0,'1',1); while($s <= $u*8){print "$u " if $s == $u; ($p,$o)=
%o (Perl) (1,0); $q=($t =~ /^(7*)/ && length $1); $r=length($t)+1; ++$o, $p *= 8 while
%o (Perl) $o+1 <= $q && $p*$r*8 <= abs($u-$s); $u += $p; s/^(7*)(.)?/(0 x length($1))
%o (Perl) .($2+1)/e, $s += tr/2/2/*$p + $o*$p/8 for substr $t,$o } print "\n"
%Y Cf. A014778.
%K nonn,base,fini,full
%O 1,1
%A _Olivier Gérard_
%E More terms and Perl program from _Hugo van der Sanden_
%E Comment from Hugo van der Sanden: Program terminates at n = 2.94239143846251e+56, when 2.37843307942386e+57 2's have been seen; since s > 8n and n > 8^8 at this point, it is not possible for n ever again to catch up with the sum (given s(8^n) = n 8^{n-1}).
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