%I #30 Sep 09 2024 09:34:41
%S 1,2,3,5,6,9,10,12,14,15,16,18,30,90,96,110,197,210,270,330,390,410,
%T 630,810,930,959,990,1110,1170,1210,1230,1470,1710,1890,1956,2310,
%U 2430,2530,2538,2710,2730,2790,2802,2922,2970,3330,3510,3519,3630,3690,4115,4245
%N Lengths k of k-digit integers of the form 1, 12, 123, 1234, ... (A057137) which are divisible by k.
%C It is easy to prove that 10*3^k, k >= 0 is always a solution.
%F {n: n|A057137(n)}. - _R. J. Mathar_, Aug 16 2021
%e 3 is a term since 123 is divisible by 3 (123 = 3*41).
%p for n from 1 to 5000 do
%p if modp(A057137(n),n) = 0 then
%p printf("%d,",n) ;
%p end if;
%p end do: # _R. J. Mathar_, Aug 16 2021
%o (Python)
%o a ="1234567890"
%o for k in range(10):
%o a = a + a
%o sol = ""
%o for n in range(1, len(a)):
%o if int(a[0:n]) % n == 0:
%o sol = sol + str(n) + ", "
%o print(sol)
%o (PARI) f(n) = 137174210*10^n\1111111111; \\ A057137
%o isok(k) = (f(k) % k) == 0; \\ _Michel Marcus_, Aug 16 2021
%Y Cf. A057137.
%K nonn,base
%O 1,2
%A _Reiner Moewald_, May 26 2021