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A350536
a(n) is the smallest proper multiple of 2n+1 which contains only odd digits, or -1 if no such multiple exists.
4
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.
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
Terms belong to A014261.
Sequence in context: A058039 A371349 A013581 * A133997 A057909 A228916
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Jan 04 2022
EXTENSIONS
More terms from Michel Marcus, Jan 04 2022
STATUS
approved