| Note that n must be a power of 2 (cf. A138083).
Similar to Fermat primes (A019434), and for the same reasons we expect this sequence to be finite as well.
The numbers (3^n + 1)/2 are strong-probable-primes to base 3, so don't test with that base. - Don Reble, Jun 15 2010
Contribution from Paul Bourdelais, Oct 13 2010: (Start)
Terms in sequence factored to 1E18:
(3^(2^21)+1)/2 has factors: 155189249
(3^(2^22)+1)/2 is composite: RES64: [A158D7ED3E1CC427](425462.3620s+0.3212s)
(3^(2^23)+1)/2 is composite: RES64: [B0F07A3D55C5082A] (3424080.8700s+1.7464s)
(3^(2^24)+1)/2 has factors: 0
(3^(2^25)+1)/2 has factors: 0
(3^(2^26)+1)/2 has factors: 3221225473
(3^(2^27)+1)/2 has factors: 0
(3^(2^28)+1)/2 has factors: 12348030977
(3^(2^29)+1)/2 has factors: 77309411329
(3^(2^30)+1)/2 has factors: 0
(3^(2^31)+1)/2 has factors: 4638564679681
(3^(2^32)+1)/2 has factors: 206158430209
(3^(2^33)+1)/2 has factors: 0
(3^(2^34)+1)/2 has factors: 50474455662593
(3^(2^35)+1)/2 has factors: 0
(3^(2^36)+1)/2 has factors: 911220261519361
(3^(2^37)+1)/2 has factors: 0
(3^(2^38)+1)/2 has factors: 6597069766657
(3^(2^39)+1)/2 has factors: 46179488366593
(3^(2^40)+1)/2 has factors: 0
(3^(2^41)+1)/2 has factors: 0
(3^(2^42)+1)/2 has factors: 0
(3^(2^43)+1)/2 has factors: 0
(3^(2^44)+1)/2 has factors: 15586676835352577
(3^(2^45)+1)/2 has factors: 16044073672507393
(3^(2^46)+1)/2 has factors: 0
(3^(2^47)+1)/2 has factors: 0
(3^(2^48)+1)/2 has factors: 0
(3^(2^49)+1)/2 has factors: 7881299347898369
(3^(2^50)+1)/2 has factors: 0
(3^(2^51)+1)/2 has factors: 891712726219358209
(3^(2^52)+1)/2 has factors: 0
(3^(2^53)+1)/2 has factors: 0
(3^(2^54)+1)/2 has factors: 180143985094819841
(End)
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