%I #63 Nov 12 2024 21:08:46
%S 1,0,1,0,1,0,1,3,2,0,6,1,3,0,2,2,1,1,2,0,14,5,1,0,1,2,5,5,2,1,4,1,1,0,
%T 2,0
%N Least k>=0 such that both 3n*2^k+1 and 3n*2^k-1 are primes, or -1 if no such k exists.
%C If n*2^k+1 and n*2^k-1 are both primes, then n must be divisible by 3.
%C 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}).
%C Conjecture: if n < 79, then a(n) >= 0.
%C 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, ...}
%C a(37), if it exists, is > 160000.
%H Ray Ballinger and Wilfrid Keller, <a href="http://www.prothsearch.com/riesel1.html">Primes of the form k*2^n+1 for k up to 300</a>
%H Ray Ballinger and Wilfrid Keller, <a href="http://www.prothsearch.com/riesel2.html">Primes of the form k*2^n-1 for k up to 300</a>
%H Eric Chen, <a href="/A253176/a253176.txt">Table of n, a(n) for n = 1..1000 (-1 if this term is unknown or does not exist)</a>
%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_049.htm">Problem 49. Sierpinski-like numbers</a>
%F If a(n) > 0, then a(2n) = a(n) - 1.
%e 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.
%t Table[k = 0; While[! PrimeQ[3*n*2^k + 1] || ! PrimeQ[3*n*2^k - 1], k++]; k, {n, 60}]
%o (PARI) a(n) = for(k=0, 2^24, if(ispseudoprime(3*n*2^k+1) && ispseudoprime(3*n*2^k-1), return(k)))
%Y Cf. A040076, A040081, A194591.
%K sign,more,hard
%O 1,8
%A _Eric Chen_, Mar 16 2015