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a(n) = (2^p^3 - 1)/(2^p^2 - 1) where p = n-th prime.
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%I #14 Dec 24 2022 04:55:01

%S 17,262657,1267650638007162390353805312001,

%T 31828687130226401637050536789795514059715404495050094614691019248562308626412218127220737

%N a(n) = (2^p^3 - 1)/(2^p^2 - 1) where p = n-th prime.

%t Table[(2^Prime[n]^3 - 1)/(2^Prime[n]^2 - 1),{n,4}] (* _Stefano Spezia_, Dec 23 2022 *)

%Y Cf. A051154, A051155, A051156.

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_