A344633 Lengths k of k-digit integers of the form 1, 12, 123, 1234, ... (A057137) which are divisible by k.

1, 2, 3, 5, 6, 9, 10, 12, 14, 15, 16, 18, 30, 90, 96, 110, 197, 210, 270, 330, 390, 410, 630, 810, 930, 959, 990, 1110, 1170, 1210, 1230, 1470, 1710, 1890, 1956, 2310, 2430, 2530, 2538, 2710, 2730, 2790, 2802, 2922, 2970, 3330, 3510, 3519, 3630, 3690, 4115, 4245

Offset: 1

Author

Reiner Moewald, May 26 2021

Keywords

• nonn
• base

Comments

It is easy to prove that 10*3^k, k >= 0 is always a solution.

Examples

			3 is a term since 123 is divisible by 3 (123 = 3*41).
		

Crossrefs

Cf. A057137.

Programs

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

PARI

f(n) = 137174210*10^n\1111111111; \\ A057137
isok(k) = (f(k) % k) == 0; \\ Michel Marcus, Aug 16 2021

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)

Formula

{n: n|A057137(n)}. - R. J. Mathar, Aug 16 2021