%I #50 Jul 30 2021 13:24:36

%S 1,37,12345679,4115226337448559670781893,

%T 1371742112482853223593964334705075445816186556927297668038408779149519890260631

%N Ratios R(k)/k for which R(k) / k is an integer, where R(k) = A002275(k) is a repunit.

%C This is the sequence where fractions of repunits (A002275) and their number of digits in base 10 R(k) / k are integers, where k is A014950(n). This happens for all k of the form k=3^m; this is true because R(3k) / R(k) = 10^(2k) + 10^n*k + 1, which is divisible by 3. Therefore R(3^m) is divisible by 3^m by induction on m. There are additional solutions in A014950.

%F a(n) = A002275(A014950(n))/A014950(n).

%e For n = 2, a(2) = 111/3 = 37. For n = 3, a(3) = 111111111/9 = 12345679.

%t s = Join[{1}, Select[Range[3, 81, 6], PowerMod[10, #, #] == 1 &]]; Table[(10^n - 1)/(9*n), {n, s}] (* _Amiram Eldar_, Jun 20 2021 *)

%o (Python) [(10**n-1)//(9*n) for n in range(1, 300) if not (10**n-1)//9 % n]

%Y Cf. A002275, A014950.

%K base,nonn

%O 1,2

%A _Thomas T. Burgess_, Jun 20 2021