A350537 Smallest number m > 1 such that (2n+1)*m = A350536(n) contains only odd digits.

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, 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, 5, 3, 3, 11, 11, 3, 11, 15, 3, 3, 13, 7, 7, 11, 5, 11, 5, 13, 5, 9, 5, 47, 5

Offset: 0

Author

Bernard Schott, Jan 12 2022

Keywords

• nonn
• base

Comments

Record values of a(n) are 3, 5, 11, 15, 17, 21, 35, 101, 155, ...

Examples

			The smallest proper multiple of 21 = 2*10+1 with only odd digits is A350536(10) = 315, as 315 = 21 * 15, a(10) = 15.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A078221, A296009, A350536, A350538, A350697.

Programs

Mathematica

Table[m=2;While[Or@@EvenQ[IntegerDigits[(2n+1)*++m]]];m,{n,0,79}] (* Giorgos Kalogeropoulos, Jan 12 2022 *)

PARI

isok(k) = my(d=digits(k)); #d == #select(x->((x%2)==1), d);
a(n) = my(k=6*n+3); while (!isok(k), k+=4*n+2); k/(2*n+1); \\ Michel Marcus, Jan 12 2022

Formula

a(n) = A350536(n) / (2n+1).

Extensions

More terms from Michel Marcus, Jan 12 2022