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%I #14 Mar 23 2018 01:58:49
%S 1,2,3,10,11,12,13,14,15,4,17,18,19,5,6,20,21,22,23,24,7,25,26,27,28,
%T 29,8,9,30,31,32,33,100,101,34,103,104,35,106,107,36,109,110,37,112,
%U 113,38,115,116,39,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136
%N a(1) = 1. For n > 1, a(n) is the smallest positive integer x not already in the sequence such that the product of x and its initial digit is minimal and strictly larger than the same product for any previous term.
%C We see in the Example section that P is the smallest possible product strictly bigger than the previous one and not leading to a contradiction.
%H Jean-Marc Falcoz, <a href="/A301382/b301382.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) * [the first digit of a(1)] = 1 * 1 = P = 1
%e a(2) * [the first digit of a(2)] = 2 * 2 = P = 4
%e a(3) * [the first digit of a(3)] = 3 * 3 = P = 9
%e a(4) * [the first digit of a(4)] = 10 * 1 = P = 10
%e a(5) * [the first digit of a(5)] = 11 * 1 = P = 11
%e a(6) * [the first digit of a(6)] = 12 * 1 = P = 12
%e a(7) * [the first digit of a(7)] = 13 * 1 = P = 13
%e a(8) * [the first digit of a(8)] = 14 * 1 = P = 14
%e a(9) * [the first digit of a(9)] = 15 * 1 = P = 15
%e a(10) * [the first digit of a(10)] = 4 * 4 = P = 16
%e a(11) * [the first digit of a(11)] = 17 * 1 = P = 17
%e Etc.
%o (PARI) p = vector(136, k, oo); for (n=1, #p, x = n*digits(n)[1]; if (x<=#p, p[x] = min(p[x], n))); for (k=1, #p, if (p[k] != oo, print1 (p[k] ", "))) \\ _Rémy Sigrist_, Mar 22 2018
%K nonn,base
%O 1,2
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Mar 20 2018
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