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A175768
Primes of the form k * b^b + 1, with b > 1.
4
5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 163, 173, 181, 193, 197, 229, 233, 241, 257, 269, 271, 277, 281, 293, 313, 317, 337, 349, 353, 373, 379, 389, 397, 401, 409, 421, 433, 449, 457, 461, 487, 509, 521, 541, 557, 569, 577, 593, 601, 613
OFFSET
1,1
COMMENTS
Without the restriction on b, the sequence would be identical to A000040.
LINKS
EXAMPLE
For a(3), 4 * 2^2 + 1 = 17, which is prime.
From Seiichi Manyama, Mar 27 2018: (Start)
n | a(n)
---+----------------------------------
1 | 5 = 1 * 2^2 + 1.
2 | 13 = 3 * 2^2 + 1.
3 | 17 = 4 * 2^2 + 1.
4 | 29 = 7 * 2^2 + 1.
5 | 37 = 9 * 2^2 + 1.
6 | 41 = 10 * 2^2 + 1.
7 | 53 = 13 * 2^2 + 1.
8 | 61 = 15 * 2^2 + 1.
9 | 73 = 18 * 2^2 + 1.
10 | 89 = 22 * 2^2 + 1.
11 | 97 = 24 * 2^2 + 1.
12 | 101 = 25 * 2^2 + 1.
13 | 109 = 27 * 2^2 + 1 = 4 * 3^3 + 1. (End)
MATHEMATICA
Take[ Select[ Union@ Flatten@ Table[ k*b^b + 1, {b, 2, 20}, {k, 148}], PrimeQ], 55] (* Robert G. Wilson v, Sep 01 2010 *)
PROG
(PARI) isA175768(n)=if(!isprime(n), return(0)); if(n%4==1||n%27==1, return(1)); forprime(b=5, log(n)/log(7), if(n%(b^b)==1, return(1))); 0 \\ Charles R Greathouse IV, Sep 02 2010
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Kevin Batista (kevin762401(AT)yahoo.com), Sep 01 2010
EXTENSIONS
Corrected and edited by Charles R Greathouse IV, Sep 02 2010
STATUS
approved