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%I #57 Jun 04 2023 08:56:10
%S 1,17,193,619,2016,91958,91958,8186278,45392361,977982331,26450915298,
%T 91600221212,196425900073,14810317269038,44430951807114,
%U 626642721222487,626642721222487,102882886570917135,874191214492184404,3830977578643912683,86801197487071715103
%N Least k > 0 such that the first n digits of 2^k and 3^k are identical.
%H Zhao Hui Du, <a href="/A359698/b359698.txt">Table of n, a(n) for n = 0..1000</a>
%e n k = a(n) 1st n digits
%e -- ----------- -------------
%e 0 1
%e 1 17 1...
%e 2 193 12...
%e 3 619 217...
%e 4 2016 7524...
%e 5 91958 13071...
%e 6 91958 130719...
%e 7 8186278 1701547...
%e 8 45392361 17179395...
%e 9 977982331 725070476...
%e 10 26450915298 2919267309...
%e a(3) = 619 because 2^619 = 2.175...*10^186 and 3^619 = 2.177...*10^295 both begin with the same three digits (in base ten), and this is not true of any smaller positive integer than 619.
%e a(0) = 1 because 2^1 and 3^1 share zero leading digits.
%Y Cf. A000079, A000244, A088995.
%K base,nonn
%O 0,2
%A _Keith F. Lynch_, May 20 2023
%E a(11)-a(20) from _Jon E. Schoenfield_, May 21 2023