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%I #25 Jan 22 2022 08:40:39
%S 3,3,3,5,11,3,3,5,3,3,15,5,3,5,11,3,3,5,3,3,13,13,3,11,11,3,3,13,3,3,
%T 13,5,3,5,11,5,7,5,7,5,17,11,7,11,11,21,15,21,35,101,11,5,3,5,11,3,3,
%U 5,3,3,11,11,3,11,15,3,3,13,7,7,11,5,11,5,13,5,9,5,47,5
%N Smallest number m > 1 such that (2n+1)*m = A350536(n) contains only odd digits.
%C Record values of a(n) are 3, 5, 11, 15, 17, 21, 35, 101, 155, ...
%H Michael De Vlieger, <a href="/A350537/b350537.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A350536(n) / (2n+1).
%e The smallest proper multiple of 21 = 2*10+1 with only odd digits is A350536(10) = 315, as 315 = 21 * 15, a(10) = 15.
%t Table[m=2;While[Or@@EvenQ[IntegerDigits[(2n+1)*++m]]];m,{n,0,79}] (* _Giorgos Kalogeropoulos_, Jan 12 2022 *)
%o (PARI) isok(k) = my(d=digits(k)); #d == #select(x->((x%2)==1), d);
%o a(n) = my(k=6*n+3); while (!isok(k), k+=4*n+2); k/(2*n+1); \\ _Michel Marcus_, Jan 12 2022
%Y Cf. A078221, A296009, A350536, A350538, A350697.
%K nonn,base
%O 0,1
%A _Bernard Schott_, Jan 12 2022
%E More terms from _Michel Marcus_, Jan 12 2022
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