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%I #18 Sep 24 2019 02:41:37
%S 1,1,1,1,1,1,1,1,1,1,15,5,3,5,11,1,1,1,1,1,13,13,3,11,11,1,1,1,1,1,13,
%T 5,3,5,11,1,1,1,1,1,17,11,7,11,11,1,1,1,1,1,11,5,3,5,11,1,1,1,1,1,11,
%U 11,3,11,15,1,1,1,1,1,11,5,11,5,13
%N Smallest number m such that (2n-1)*m has only odd digits.
%C Record values:
%C 1 * 1 = 1
%C 21 * 15 = 315
%C 81 * 17 = 1377
%C 167 * 19 = 3173
%C 169 * 33 = 5577
%C 201 * 155 = 31155
%C 633 * 283 = 179139
%C 1011 * 743 = 751173
%C 1101 * 833 = 917133
%C 2001 * 1555 = 3111555
%C 9091 * 4309 = 39173119
%C 9901 * 32231 = 319119131
%C 91001 * 34193 = 3111597193
%C 100011 * 37927 = 3793117197
%C 101001 * 58553 = 5913911553
%C 200001 * 155555 = 31111155555
%C 909091 * 431109 = 391917311919
%C 990001 * 12121113 = 11999913991113
%C 999001 * 31222311 = 31191119911311
%C ... (above are exhaustive)
%C 99990001 * 31122223111 = 3111911119991113111 (verified smallest)
%C 9999900001 * 31112222231111 = 311119111119999111131111 (not verified smallest)
%H Yang Haoran, <a href="/A296009/b296009.txt">Table of n, a(n) for n = 1..10001</a>
%H Yang Haoran, <a href="/A296009/a296009.txt">solver program for specified 2n-1, using faster algorithm than the brute-force approach</a>
%F a(n) = A061808(n)/(2n-1).
%e For n = 11, 2n-1 = 21, 21*15 = 315 which has all odd digits.
%e For m = 1 to 14, n*m listed are 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294, all of which contains at least one even digit.
%t f[n_] := Block[{m = 1, nn = 2n -1, od = {1, 3, 5, 7, 9}}, While[ Union@ Join[od, IntegerDigits[m*nn]] != od, m += 2]; m]; Array[f, 75] (* _Robert G. Wilson v_, Dec 05 2017 *)
%o (PARI) isok(n, m) = {my(d = digits((2*n-1)*m)); #select(x->((x%2)==0), d) == 0;}
%o a(n) = {my(m=1); while (!isok(n, m), m++); m;} \\ _Michel Marcus_, Sep 24 2019
%Y Cf. A061808, A216473.
%K base,nonn,look
%O 1,11
%A _Yang Haoran_, Dec 02 2017
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