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%I #30 Aug 02 2018 08:39:53
%S 1,1,1,1,1,1,1,1,4,1,3,16,1,11,6,37,1,9,120,9,1,2,67,16,1,26,103,12,
%T 60,1,239,4,40,2,44,174,33,1,3,260,114,1,161,70,1,3,2,3,50,45,472,228,
%U 183,66,37,7,122,235,68,102,294,8,13,1,40,62,143,1,61,7
%N Smallest positive number b such that ((b+1)^prime(n) + b^prime(n))/(2*b + 1) is prime, or 0 if no such b exists.
%C Conjecture: a(n) > 0 for every n > 1.
%C Records: 1, 4, 16, 37, 120, 239, 260, 472, 917, 1539, 6633, 7050, 12818, ..., which occur at n = 2, 10, 13, 17, 20, 32, 41, 52, 72, 128, 171, 290, 309, ... - _Robert G. Wilson v_, Jun 16 2018
%H Robert G. Wilson v, <a href="/A301510/b301510.txt">Table of n, a(n) for n = 2..345</a>
%H Richard Fischer, <a href="http://www.fermatquotient.com/PrimSerien/PrimPotP.txt">Primzahlen mit der Form [(B+1)^N+B^N]/(2*B+1)</a>
%H Henri Lifchitz & Renaud Lifchitz, <a href="http://www.primenumbers.net/prptop/searchform.php?form=%28a%5En%2Bb%5En%29%2Fc&action=Search">Search for: (a^n+b^n)/c</a>
%F a(n) = A250201(2*prime(n)) - 1 for n >= 2. - _Eric Chen_, Jun 06 2018
%e a(10) = 4 because (5^29 + 4^29)/9 = 2149818248341 is prime and (2^29 + 1^29)/3, (3^29 + 2^29)/5 and (4^29 + 3^29)/7 are all composite.
%t Table[p = Prime[n]; k = 1; While[q = ((b+1)^n+b^n)/(2*b+1); ! PrimeQ[q], k++]; k, {n, 200}]
%t f[n_] := Block[{b = 1, p = Prime@ n}, While[! PrimeQ[((b +1)^p + b^p)/(2b +1)], b++]; b]; Array[f, 70, 2] (* _Robert G. Wilson v_, Jun 13 2018 *)
%o (PARI) for(n=2, 200, b=0; until(isprime((((b+1)^prime(n)+b^prime(n))/(2*b+1))), b++); print1(b,", ")) \\ corrected by _Eric Chen_, Jun 06 2018
%Y Numbers n such that ((b+1)^n + b^n)/(2*b + 1) is prime for b = 1 to 18: A000978, A057469, A128066, A128335, A128336, A187805, A181141, A187819, A217095, A185239, A213216, A225097, A224984, A221637, A227170, A228573, A227171, A225818.
%Y Cf. A058013, A103794, A222119, A247244, A250201.
%K nonn
%O 2,9
%A _Tim Johannes Ohrtmann_, Mar 22 2018
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