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A288160
a(n) = smallest k such that (6*k*n-3)*2^n-1 is prime, or 0 if no such prime exists.
1
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
1,2
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.
CROSSREFS
Cf. A285808.
Sequence in context: A160096 A029446 A358479 * A275332 A029442 A125917
KEYWORD
nonn
AUTHOR
Pierre CAMI, Jun 19 2017
STATUS
approved