A301510 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.

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, 60, 1, 239, 4, 40, 2, 44, 174, 33, 1, 3, 260, 114, 1, 161, 70, 1, 3, 2, 3, 50, 45, 472, 228, 183, 66, 37, 7, 122, 235, 68, 102, 294, 8, 13, 1, 40, 62, 143, 1, 61, 7

Offset

2,9

Comments

Conjecture: a(n) > 0 for every n > 1.

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

Links

Robert G. Wilson v, Table of n, a(n) for n = 2..345

Richard Fischer, Primzahlen mit der Form [(B+1)^N+B^N]/(2*B+1)

Henri Lifchitz & Renaud Lifchitz, Search for: (a^n+b^n)/c

Formula

a(n) = A250201(2*prime(n)) - 1 for n >= 2. - Eric Chen, Jun 06 2018

Example

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.

Mathematica

Table[p = Prime[n]; k = 1; While[q = ((b+1)^n+b^n)/(2*b+1); ! PrimeQ[q], k++]; k, {n, 200}]
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 *)

Prog

(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

Crossrefs

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.

Cf. A058013, A103794, A222119, A247244, A250201.

Sequence in context: A162516 A336693 A193793 * A085471 A064221 A378592

Adjacent sequences: A301507 A301508 A301509 * A301511 A301512 A301513

Keyword

nonn

Author

Tim Johannes Ohrtmann, Mar 22 2018

Status

approved