OFFSET
1,2
COMMENTS
It is easy to prove that 10*3^k, k >= 0 is always a solution.
FORMULA
{n: n|A057137(n)}. - R. J. Mathar, Aug 16 2021
EXAMPLE
3 is a term since 123 is divisible by 3 (123 = 3*41).
MAPLE
for n from 1 to 5000 do
if modp(A057137(n), n) = 0 then
printf("%d, ", n) ;
end if;
end do: # R. J. Mathar, Aug 16 2021
PROG
(Python)
a ="1234567890"
for k in range(10):
a = a + a
sol = ""
for n in range(1, len(a)):
if int(a[0:n]) % n == 0:
sol = sol + str(n) + ", "
print(sol)
(PARI) f(n) = 137174210*10^n\1111111111; \\ A057137
isok(k) = (f(k) % k) == 0; \\ Michel Marcus, Aug 16 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Reiner Moewald, May 26 2021
STATUS
approved