A238797 Smallest k such that 2^k - (2*n+1) and (2*n+1)*2^k - 1 are both prime, k <= 2*n+1, or 0 if no such k exists.
0, 3, 4, 0, 0, 0, 0, 5, 6, 5, 7, 6, 9, 5, 0, 7, 6, 6, 0, 0, 10, 0, 6, 0, 7, 9, 6, 7, 8, 0, 17, 8, 0, 0, 7, 0, 0, 18, 0, 0, 0, 8, 0, 10, 8, 9, 18, 0, 0, 7, 0, 0, 8, 12, 0, 7, 0, 11, 16, 0, 21, 0, 0, 0, 8, 14, 0, 0, 18, 9, 10, 8, 77, 0, 0, 0, 12, 8, 0, 11, 18, 0
Offset: 0
Keywords
Examples
a(1) = 3 because 2^3 - (2*1+1) = 5 and (2*1+1)*2^3 - 1 = 23 are both prime, 3 = 2*1+1, a(2) = 4 because 2^4 - (2*2+1) = 11 and (2*2+1)*2^4 - 1 = 79 are both prime, 4 < 2*2+1 = 5.
Links
- Giovanni Resta, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Mathematica
a[n_] := Catch@ Block[{k = 1}, While[k <= 2*n+1, If[2^k - (2*n + 1) > 0 && PrimeQ[2^k - (2*n+1)] && PrimeQ[(2*n + 1)*2^k-1], Throw@k]; k++]; 0]; a/@ Range[0, 80] (* Giovanni Resta, Mar 15 2014 *)
Extensions
a(0), a(19), a(20) corrected by Giovanni Resta, Mar 13 2014
Comments