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%I #36 Mar 16 2019 17:14:05
%S 2,2,14,4,2,14,126,8,1022,2,6,28,126,126,14,16,131070,1022,524286,4,
%T 126,6,8388606,56,4,126,268435454,252,536870910,14,65534,32,126,
%U 131070,126,2044,14,524286,126,8,62,126,4194302,12,1022,8388606,140737488355326,112,8796093022206
%N Any positive integer n has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983). This multiple is regarded as a binary number and a(n) is its conversion to base 10.
%C For any odd number m not divisible by 5 (A045572), Euler's theorem (lcm(9*m,10) = 1, so 10^phi(9*m) == 1 (mod 9*m); i.e., 9*m | 10^d - 1 = 9*R_d with d = phi(9*m)) guarantees that the repunit R_d is always some multiple of m.
%C The numbers of the form 2^i*5^j with i, j >= 0 (A003592) clearly have a multiple equal to 10^r, for r = max(i,j).
%C These multiples of n end in a string of one or more 0's, so all the terms of this sequence are even.
%C The powers 2^k are fixed points of this sequence: the smallest multiple of 2^k consisting of a succession of 1's followed by a succession of 0's is 10^k, and 10^k in base 2 is 2^k in base 10.
%H Makoto Kamada, <a href="https://stdkmd.net/nrr/repunit/">Factorization of 11...11 (Repunit)</a>.
%e The smallest multiple of the integer 7 consisting of a succession of 1's followed by a succession of 0's is 1111110, and 1111110_2 = 126_10, so a(7) = 126. This is also the case for n=13, 14, 21, 26, 33, 35, 37, ...
%Y Cf. A000079, A052983, A045572, A099679, A002275, A007088, A276349, A003592, A011557, A276348.
%K nonn,base
%O 1,1
%A _Bernard Schott_, Feb 22 2019
%E More terms from _Michel Marcus_, Feb 28 2019
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