A350536 - OEIS

%I #45 Jan 12 2022 01:02:06

%S 3,9,15,35,99,33,39,75,51,57,315,115,75,135,319,93,99,175,111,117,533,

%T 559,135,517,539,153,159,715,171,177,793,315,195,335,759,355,511,375,

%U 539,395,1377,913,595,957,979,1911,1395,1995,3395,9999,1111,515,315,535,1199,333

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

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

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

%H Chai Wah Wu, <a href="/A350536/b350536.txt">Table of n, a(n) for n = 0..10000</a>

%H International Mathematical Talent Search, <a href="https://www2.cms.math.ca/Competitions/IMTS/imts2.html">Problem 1/2</a>, Round 2.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical Competitions</a>.

%e a(10) = 315 = 21 * 15 is the smallest multiple of 21 which contains only odd digits.

%e 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.

%t 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 *)

%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; \\ _Michel Marcus_, Jan 04 2022

%o (Python)

%o from itertools import product, count

%o def A350536(n):

%o     m = 2*n+1

%o     for l in count(len(str(m))):

%o         for s in product('13579',repeat=l):

%o             k = int(''.join(s))

%o             if k > m and k % m == 0:

%o                 return k # _Chai Wah Wu_, Jan 11 2022

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

%Y Terms belong to A014261.

%K nonn,base

%O 0,1

%A _Bernard Schott_, Jan 04 2022

%E More terms from _Michel Marcus_, Jan 04 2022