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%I #15 Mar 23 2019 05:30:51
%S 1,31,119,185,511,2341,9671,7721,67449,364579,513334,639227,6250772,
%T 30377688,82011443,78927181
%N a(n) is the least exponent e such that 3^e has exactly n consecutive 3's in its decimal representation.
%C No more terms < 10^8. - _Bert Dobbelaere_, Mar 20 2019
%e a(3)=119 because 3^119 (i.e., 599003433304810403471059943169868346577158542512617035467) is the smallest power of 3 to contain a run of 3 consecutive threes in its decimal form.
%t a = ""; Do[ a = StringJoin[a, "3"]; b = StringJoin[a, "3"]; k = 1; While[ StringPosition[ ToString[3^k], a] == {} || StringPosition[ ToString[3^k], b] != {}, k++ ]; Print[k], {n, 1, 10} ]
%Y Cf. A195269, A131552, A131551, A131549, A131548, A131547, A131546, A131545, A131544.
%K more,nonn,base
%O 1,2
%A _Shyam Sunder Gupta_, Aug 26 2007
%E a(11)-a(13) from _Lars Blomberg_, Feb 02 2013
%E a(14)-a(16) from _Bert Dobbelaere_, Mar 20 2019
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