%I #51 Feb 28 2020 00:03:53
%S 1,91,8190090090909099099181
%N a(n) = R_(prime(n)#) / Product_{j=1..n} R_(prime(j)), where prime(n)# is the n-th primorial number A002110(n) and R_k = (10^k - 1)/9.
%C a(4) has 196 digits.
%C The numbers R_k = 1, 11, 111, ... are sometimes called "Rep-units" or "repunits". The octal versions of a(1) through a(4) may be obtained from the decimal versions by replacing each 6 with a 4, each 7 with a 5, each 8 with a 6, and each 9 with a 7. Similar relations exist for other bases.
%D Author?, "The Ultimate Number Series Challenge", Vidya, Oct 1988, p. 9.
%H Michel Marcus, <a href="/A323060/b323060.txt">Table of n, a(n) for n = 1..4</a>
%H Patrick A. Thomas, <a href="/A323060/a323060_3.txt">Term a(5) and information on a(6)</a>
%e R_30 / (11*111*11111) = 8190090090909099099181.
%t f[n_] := (10^n - 1)/9; Array[f[Product[Prime@ i, {i, #}]]/Product[f@ Prime@ j, {j, #}] &, 3] (* _Michael De Vlieger_, Jan 19 2019 *)
%o (PARI) R(n) = (10^n-1)/9; \\ A002275
%o primo(n) = prod(i=1, n, prime(i)); \\ A002110
%o a(n) = R(primo(n))/prod(j=1, n, R(prime(j))); \\ _Michel Marcus_, Jan 21 2019
%Y Cf. A002110, A002275, A031974.
%K nonn,easy
%O 1,2
%A _Patrick A. Thomas_, Jan 19 2019