A350536 a(n) is the smallest proper multiple of 2n+1 which contains only odd digits, or -1 if no such multiple exists.

3, 9, 15, 35, 99, 33, 39, 75, 51, 57, 315, 115, 75, 135, 319, 93, 99, 175, 111, 117, 533, 559, 135, 517, 539, 153, 159, 715, 171, 177, 793, 315, 195, 335, 759, 355, 511, 375, 539, 395, 1377, 913, 595, 957, 979, 1911, 1395, 1995, 3395, 9999, 1111, 515, 315, 535, 1199, 333

Offset: 0

Author

Bernard Schott, Jan 04 2022

Keywords

• nonn
• base

Comments

Generalization of the problem 1/2 of International Mathematical Talent Search, round 2 (see link and 2nd example).

If the escape clause is used, it will be necessarily for terms coming from n = 12 + 25*k, k >= 0.

Examples

			a(10) = 315 = 21 * 15 is the smallest multiple of 21 which contains only odd digits.
a(4998) = 33339995 = 9997 * 3335 is the smallest multiple of 9997 which contains only odd digits, so this is the answer to the IMTS problem.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000
• International Mathematical Talent Search, Problem 1/2, Round 2.
• Index to sequences related to Olympiads and other Mathematical Competitions.

Crossrefs

Cf. A061810, A061829, A078221, A296009, A350538.

Terms belong to A014261.

Programs

Mathematica

a[n_] := Module[{m = 2*n + 1, k}, k = 3*m; While[!AllTrue[IntegerDigits[k], OddQ], k += 2*m]; k]; Array[a, 50, 0] (* Amiram Eldar, Jan 04 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; \\ Michel Marcus, Jan 04 2022

Python

from itertools import product, count
def A350536(n):
    m = 2*n+1
    for l in count(len(str(m))):
        for s in product('13579',repeat=l):
            k = int(''.join(s))
            if k > m and k % m == 0:
                return k # Chai Wah Wu, Jan 11 2022

Extensions

More terms from Michel Marcus, Jan 04 2022