A253176 Least k>=0 such that both 3n*2^k+1 and 3n*2^k-1 are primes, or -1 if no such k exists.

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, 2, 0

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

Table of n, a(n) for n=1..36.

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

Eric Chen, Table of n, a(n) for n = 1..1000 (-1 if this term is unknown or does not exist)

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

Cf. A040076, A040081, A194591.

Sequence in context: A359843 A364361 A338022 * A079408 A114376 A216395

Adjacent sequences: A253173 A253174 A253175 * A253177 A253178 A253179

Keyword

sign,more,hard

Author

Eric Chen, Mar 16 2015

Status

approved