A125958 Least number k > 0 such that (2^k + (2n-1)^k)/(2n+1) is prime.

3, 3, 3, 5, 3, 3, 7, 3, 5, 5, 11, 3, 19, 11, 3, 229, 47, 5, 257, 3, 19, 31, 17, 11, 13, 3, 3, 5, 5, 59, 23, 3, 3, 7, 79, 3, 3373, 3, 3, 7, 13, 7, 7, 3527, 593, 19, 3, 3, 13, 13, 11, 19, 41, 3, 7, 109, 3, 227, 13, 5, 5, 3, 239, 5, 3251, 3, 1237, 3, 7, 31, 3, 7

Offset: 1

Author

Alexander Adamchuk, Feb 06 2007

Keywords

• nonn
• hard

Comments

All terms are odd primes.

a(38),...,a(43) = {3,3,7,13,7,7}.

a(46),...,a(64) = {19,3,3,13,13,11,19,41,3,7,109,11,227,13,5,5,3,239,5}.

a(66) = 3. a(68),...,a(72) = {3,7,31,3,7}.

a(74),...,a(92) = {3,5,19,17,3,83,3,3,19,19,11,11,61,3,7,7,3,41,29}.

a(94) = 5. a(97),a(98) = {19,7}. a(100) = 31.

a(n) is currently unknown for n = {37,44,45,65,67,73,93,95,96,99,...}.

From Kevin P. Thompson, May 18 2022: (Start)

All known terms from n=1..100 correspond to proven primes.

a(96) > 10250.

a(99) > 10250. (End)

Presuming every prime is seen at least once, one can specifically seek those with fixed k. Doing this, a(174) = 37, a(368) = 43 for example. - Bill McEachen, Aug 26 2024

Examples

			For n=4, the expression (2^k + (2n-1)^k)/(2n+1) takes on values 1, 53/9, 39, 2417/9, and 1871 for k=1..5. Since 1871 is the first prime number to occur, a(4) = 5.
		

Links

• Kevin P. Thompson, Table of n, a(n) for n = 1..95

Crossrefs

Cf. A000978 ((2^n + 1)/3 is prime), A057469 ((2^n + 3^n)/5 is prime).

Cf. A082387 ((2^n + 5^n)/7 is prime), A125955 ((2^n + 7^n)/9 is prime).

Cf. A125956 ((2^n + 9^n)/11 is prime), A125955 ((2^n + 11^n)/13 is prime).

Programs

Mathematica

Do[k = 1; While[ !PrimeQ[(2^k + (2n-1)^k)/(2n+1)], k++ ]; Print[k], {n, 100}] (* Ryan Propper, Mar 29 2007 *)

Extensions

More terms from Ryan Propper, Mar 29 2007

a(65)-a(72) from Kevin P. Thompson, May 18 2022