A256787 Smallest odd number k such that k*2^(2*n+1)+1 is a prime number.

1, 5, 3, 5, 15, 9, 5, 5, 9, 11, 11, 45, 5, 15, 23, 35, 9, 59, 15, 5, 3, 9, 35, 27, 23, 17, 51, 5, 29, 27, 53, 9, 9, 9, 23, 39, 23, 5, 29, 249, 9, 51, 5, 75, 51, 117, 29, 77, 131, 219, 221, 29, 53, 105, 321, 95, 179, 197, 101, 51, 81, 101, 11, 5, 21, 221, 53

Offset: 0

Author

Pierre CAMI, Apr 10 2015

Keywords

• nonn

Comments

As N increases, (Sum_{n=1..N} a(n))/(Sum_{n=1..N} n) appears to tend to log(2), as can be seen by plotting the first 10000 terms.

This observation is consistent with the prime number theorem as the probability that k*2^n+1 is prime is 1/(n*log(2)+log(k)) so ~ 1/(n*log(2)) as n increases, if k ~ n*log(2) then k/(n*log(2)) ~ 1.

Examples

			1*2^(2*0+1)+1=3 is prime, so a(0)=1.
1*2^(2*1+1)+1=9 and 3*2^(2*1+1)+1=25 are composite, 5*2^(2*1+1)+1=41 is prime, so a(1)=5.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 0..10000

Programs

Maple

for n from 0 to 100 do
R:= 2^(2*n+1);
for k from 1 by 2 do
   if isprime(k*R+1) then A[n]:= k; break fi
od:
od:
seq(A[n],n=0..100); # Robert Israel, Apr 24 2015

Mathematica

f[n_] := Block[{g, i, k}, g[x_, y_] := y*2^(2*x + 1) + 1; Reap@ For[i = 0, i <= n, i++, k = 1; While[Nand[PrimeQ[g[i, k]] == True, OddQ@ k], k++]; Sow@ k] // Flatten // Rest]; f@ 66 (* Michael De Vlieger, Apr 18 2015 *)

PARI

vector(100, n, n--; k=1; while(!isprime(k*2^(2*n+1)+1), k+=2); k) \\ Colin Barker, Apr 10 2015