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a(n) = 2^(2^n+n) - 1.
0

%I #10 Jun 07 2023 08:52:11

%S 1,7,63,2047,1048575,137438953471,1180591620717411303423,

%T 43556142965880123323311949751266331066367,

%U 29642774844752946028434172162224104410437116074403984394101141506025761187823615

%N a(n) = 2^(2^n+n) - 1.

%C An integer is simultaneously a Mersenne number and a Woodall number if and only if it is a member of this sequence. Hence this sequence is the intersection of A000225 and A003261.

%H Wilfrid Keller, <a href="https://doi.org/10.1090/S0025-5718-1995-1308456-3">New Cullen Primes</a>, Mathematics of Computation, Vol. 64, No. 212 (Ocober 1995), pp. 1733-1741.

%F a(n) = 2^(2^n+n)-1 = A000225(2^n+n) = A003261(2^n).

%e The fourth integer which is both a Mersenne number and a Woodall number is 2047. Hence a(3)=2047 (as the offset is zero).

%t 2^(2^#+#)-1 &/@Range[0,8]

%Y Cf. A000225, A003261, A006127.

%K easy,nonn

%O 0,2

%A _Ant King_, Feb 12 2008