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

0,1

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.

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.

Example

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.

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 *)

Prog

(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

Crossrefs

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

Terms belong to A014261.

Sequence in context: A058039 A371349 A013581 * A133997 A057909 A228916

Adjacent sequences: A350533 A350534 A350535 * A350537 A350538 A350539

Keyword

nonn,base

Author

Bernard Schott, Jan 04 2022

Extensions

More terms from Michel Marcus, Jan 04 2022

Status

approved