%I #18 Jul 26 2015 04:28:24
%S 8,17,57,177,2169,8361,131361,524649,8389137,536871753,2147484609,
%T 137438954841,2199023257233,8796093024057,140737488357537,
%U 9007199254743801,576460752303426969,2305843009213697673,147573952589676417417,2361183241434822611889
%N Numbers of the form p^2 + 2^p for p prime.
%C For any n>=3, a(n) is divisible by 3. This follows from the following simple result, combined with the fact that A061725(n), n>=3, is divisible by 3: Let r>=5 be an odd integer such that r^2 + 2 is divisible by 3. Then r^2 + 2^i is divisible by 3 for any odd integer i>=3. In particular, r^2 + 2^r is divisible by 3. This contribution was inspired by Problem of the Month - Math Central, MP98 (problem for October 2010), which asks for all primes p such that 2^p + p^2 is also a prime. - Shai Covo (green355(AT)netvision.net.il), Nov 02 2010
%H Alois P. Heinz, <a href="/A097058/b097058.txt">Table of n, a(n) for n = 1..200</a>
%e For example, the first two terms are 2^2 + 2^2 = 8, 3^2 + 2^3 = 17
%p a:= proc(n) local p; p:= ithprime(n); p^2+2^p end:
%p seq(a(n), n=1..25); # _Alois P. Heinz_, May 15 2013
%t Table[ Prime[n]^2 + 2^Prime[n], {n, 16}] (* _Robert G. Wilson v_, Sep 15 2004 *)
%t #^2+2^#&/@Prime[Range[20]] (* _Harvey P. Dale_, Jul 12 2011 *)
%o (PARI) forprime(p=2,61,print1(p^2+2^p,",")) \\ _Klaus Brockhaus_
%K nonn,easy
%O 1,1
%A _Parthasarathy Nambi_, Sep 15 2004
%E More terms from _Klaus Brockhaus_, _Ray Chandler_ and _Robert G. Wilson v_, Sep 15 2004
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