A214415 Numbers n such that prevprime(2^n) AND nextprime(2^n) = 1, where AND is the bitwise AND operator.
2, 4, 6, 8, 12, 15, 16, 23, 25, 30, 37, 53, 55, 57, 67, 75, 76, 81, 82, 84, 95, 108, 129, 132, 135, 139, 143, 155, 160, 163, 180, 181, 188, 192, 203, 204, 210, 222, 244, 263, 273, 277, 280, 287, 289, 295, 297, 308, 315, 319, 325, 330, 341, 367, 370, 393, 394, 406
Offset: 0
Examples
4 is in the sequence because (prevprime(2^4) AND nextprime(2^4)) = 13 AND 17 = 1.
Programs
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Java
import java.math.BigInteger; public class A214415 { public static void main (String[] args) { BigInteger b1 = BigInteger.valueOf(1); BigInteger b2 = BigInteger.valueOf(2); for (int n=2; ; n++) { BigInteger pwr = b1.shiftLeft(n); BigInteger pm = pwr.subtract(b1); BigInteger pp = pwr.add(b1); while (true) { if (pm.isProbablePrime(2)) { if (pm.isProbablePrime(80)) break; } pm = pm.subtract(b2); } while (true) { if (pp.isProbablePrime(2)) { if (pp.isProbablePrime(80)) break; } pp = pp.add(b2); } if (pm.and(pp).equals(b1)) { System.out.printf("%d, ",n); } } } }
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Mathematica
ba1Q[n_]:=Module[{c=2^n},BitAnd[NextPrime[c],NextPrime[c,-1]]==1]; Select[ Range[ 450],ba1Q] (* Harvey P. Dale, Dec 25 2012 *)
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PARI
{ for (n=2,1000, N = 2^n; p1 = precprime(N-1); p2 = nextprime(N+1); ba = bitand(p1, p2); if ( bitand( ba, ba-1 ) == 0, print1(n,", ")); ); } /* Joerg Arndt, Aug 16 2012 */
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Python
from itertools import islice from sympy import prevprime, nextprime def A214415_gen(): # generator of terms n, m = 2, 4 while True: if prevprime(m)&nextprime(m) == 1: yield n n += 1 m *= 2 A214415_list = list(islice(A214415_gen(),20)) # Chai Wah Wu, Oct 16 2023
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