A174057 - OEIS

%I #7 Aug 09 2017 22:15:02

%S 3,4,5,6,10,12,17,18,24,65,66,72,130,132,144,192,4097,4098,4104,4224,

%T 32770,32772,32784,32832,36864,65537,65538,65544,65664,98304,262145,

%U 262146,262152,262272,294912,1073741825,1073741826,1073741832

%N Semi-sums (means) of a Fermat prime and a Mersenne prime.

%C The subsequence of prime semi-sums (means) of a Fermat prime and a Mersenne prime begins: 3, 5, 17, 65537 = (3 + 131071)/2. _R. J. Mathar_, on the remaining primes in the half sum, searched through all sums that can be created from the existing values of the two OEIS sequences, and that the next Fermat prime is known to be > 2^(2^32) + 1. So it is safe to say that the next prime > 65537 in the half sum (if it exists) is larger than 85070591730234615865843651857942085632, because adding that huge next Fermat prime would lead to even larger numbers. Of course one could easily boost that estimate by using the b-file of A000668.

%F {(A019434(i) + A000668(j))/2}. {(((2^p)-1) + (2^(2^k)+1))/2 = 2^(p-1) + 2^((2^k)-1) for p in A000043 and k in {0,1,2,3,4}}.

%e a(1) = 3 = half of first Mersenne prime + first Fermat prime = (3+3)/2.

%e a(2) = 4 = half of first Mersenne prime + 2nd Fermat prime = (3+5)/2.

%e a(3) = 5 = half of 2nd Mersenne prime + first Fermat prime = (7+3)/2.

%e a(4) = 6 = half of 2nd Mersenne prime + 2nd Fermat prime = (7+5)/2.

%e a(5) = 10 = half of 2nd Mersenne prime + 3rd Fermat prime = (3+17)/2.

%Y Cf. A000668, A171251-A171255, A155877 Sums of three Fermat numbers, A166484 Prime sums of three Fermat numbers, A174055, A174056.

%K nonn

%O 1,1

%A _Jonathan Vos Post_, Mar 06 2010