%I #13 Jan 12 2022 10:38:53
%S 1,3,9,99,9999,99999999,9999999999999999,
%T 99999999999999999999999999999999,
%U 9999999999999999999999999999999999999999999999999999999999999999
%N a(1) = 1, a(n+1) > a(n) is the smallest multiple of a(n) using only odd digits.
%F a(n) = 10^(2^(n-3)) - 1 for n >= 3. (Proof by induction. Consider a(n)*f, L = ceiling(log(f)/log(10)), g1 = number formed by the first L digits of a(n)*f, g2 = number formed by the last L digits of a(n)*f => g1 + g2 = number formed by L 9's, if L <= 10^(2^(n-2)) + 1). - _Sascha Kurz_, Jan 04 2003
%p 1,3,seq(10^(2^(n-3))-1,n=3..11);
%o (Python)
%o def A078221(n): return 2*n-1 if n < 3 else 10**(2**(n-3)) - 1 # _Chai Wah Wu_, Jan 12 2022
%Y Cf. A078222.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Nov 22 2002
%E More terms from _Sascha Kurz_, Jan 04 2003