A288160 a(n) = smallest k such that (6*k*n-3)*2^n-1 is prime, or 0 if no such prime exists.

1, 2, 2, 2, 1, 1, 0, 1, 1, 1, 1, 4, 6, 6, 0, 15, 2, 8, 3, 1, 9, 4, 3, 14, 1, 0, 3, 0, 1, 2, 1, 3, 4, 25, 0, 1, 24, 2, 17, 22, 2, 4, 16, 2, 13, 9, 17, 17, 0, 10, 17, 3, 6, 34, 0, 1, 69, 5, 26, 8, 4, 3, 3, 8, 16, 19, 3, 5, 5, 0, 18, 8, 75, 5, 0, 1, 0, 37, 19, 14, 85, 4, 4, 47

Offset: 1

Author

Pierre CAMI, Jun 19 2017

Keywords

• nonn

Comments

For some n (6*k*n-3)*2^n-1 is composite for any k.

For n=15+20*j, n=7+21*j, n=77+110*j, n=26+156*j, n=266+342*j, n=261+812*j, n=2368+1332*j, n=477+2756*j, n=2183+3422*j and more others (6*k*n-3)*2^n-1 is always composite for any k and any j.

For n=4390+187892*j, (6*k*n-3)*2^n-1 is always divisible by one of the 82 primes between 5 and 443, 4390=10*439 and 187892=438*439.

For n=6152+596744*j, (6*k*n-3)*2^n-1 is always divisible by one of the 134 primes between 3 and 773, 6152=8*769 and 596744=768*769.

For n=11*1229+1228*1229*j, (6*n*k-3)*2^n-1 is always divisible by one of the 199 primes between 3 and 1231 except 11.

For n=27*1399+1398*1399*j, (6*n*k-3)*2^n-1 is always divisible by one of the 220 primes between 3 and 1409.

For n=5*11*1619+1618*1619*j, (6*n*k-3)*2^n-1 is always divisible by one of the 253 primes between 5 and 1621 except 11.

Links

• Pierre CAMI, Table of n, a(n) for n = 1..9000

Crossrefs

Cf. A285808.