OFFSET
1,8
COMMENTS
If n*2^k+1 and n*2^k-1 are both primes, then n must be divisible by 3.
a(79) = -1, since 237*2^k+1 or 237*2^k-1 must divisible by 5, 7, 13, 17, or 241. Similarly, a(269) = -1 (cover: {5, 7, 13, 19, 37, 73}), a(1527) = -1 (cover: {5, 7, 13, 17, 241}).
Conjecture: if n < 79, then a(n) >= 0.
a(38) - a(40) = {1, 4, 1}, a(42) - a(50) = {13, 3, 4, 1, 0, 1, 3, 44, 0}, a(52) = 1, a(54) - a(56) = {4, 2, 4}, a(58) - a(60) = {1, 12, 0}, a(n) is currently unknown for n = {37, 41, 51, 53, 57, ...}
a(37), if it exists, is > 160000.
LINKS
Ray Ballinger and Wilfrid Keller, Primes of the form k*2^n+1 for k up to 300
Ray Ballinger and Wilfrid Keller, Primes of the form k*2^n-1 for k up to 300
Carlos Rivera, Problem 49. Sierpinski-like numbers
FORMULA
If a(n) > 0, then a(2n) = a(n) - 1.
EXAMPLE
a(11) = 6 since 33*2^n+1 and 33*2^n-1 are not both primes for all 0 <= n <= 5, but they are both primes for n = 6.
MATHEMATICA
Table[k = 0; While[! PrimeQ[3*n*2^k + 1] || ! PrimeQ[3*n*2^k - 1], k++]; k, {n, 60}]
PROG
(PARI) a(n) = for(k=0, 2^24, if(ispseudoprime(3*n*2^k+1) && ispseudoprime(3*n*2^k-1), return(k)))
CROSSREFS
KEYWORD
sign,more,hard,changed
AUTHOR
Eric Chen, Mar 16 2015
STATUS
approved