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a(1) = 1, a(n+1) > a(n) is the smallest multiple of a(n) using only odd digits.
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%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