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A214415 Numbers n such that prevprime(2^n) AND nextprime(2^n) = 1, where AND is the bitwise AND operator. 2

%I

%S 2,4,6,8,12,15,16,23,25,30,37,53,55,57,67,75,76,81,82,84,95,108,129,

%T 132,135,139,143,155,160,163,180,181,188,192,203,204,210,222,244,263,

%U 273,277,280,287,289,295,297,308,315,319,325,330,341,367,370,393,394,406

%N Numbers n such that prevprime(2^n) AND nextprime(2^n) = 1, where AND is the bitwise AND operator.

%C A007053(a(n)) are indices of 1's in A175330. That is, A175330(A007053(a(n)))=1.

%C Conjecture: the sequence is infinite.

%e 4 is in the sequence because (prevprime(2^4) AND nextprime(2^4)) = 13 AND 17 = 1.

%t ba1Q[n_]:=Module[{c=2^n},BitAnd[NextPrime[c],NextPrime[c,-1]]==1]; Select[ Range[ 450],ba1Q] (* _Harvey P. Dale_, Dec 25 2012 *)

%o (Java)

%o import java.math.BigInteger;

%o public class A214415 {

%o public static void main (String[] args) {

%o BigInteger b1 = BigInteger.valueOf(1);

%o BigInteger b2 = BigInteger.valueOf(2);

%o for (int n=2; ; n++) {

%o BigInteger pwr = b1.shiftLeft(n);

%o BigInteger pm = pwr.subtract(b1);

%o BigInteger pp = pwr.add(b1);

%o while (true) {

%o if (pm.isProbablePrime(2)) {

%o if (pm.isProbablePrime(80)) break;

%o }

%o pm = pm.subtract(b2);

%o }

%o while (true) {

%o if (pp.isProbablePrime(2)) {

%o if (pp.isProbablePrime(80)) break;

%o }

%o pp = pp.add(b2);

%o }

%o if (pm.and(pp).equals(b1)) {

%o System.out.printf("%d, ",n);

%o }

%o }

%o }

%o }

%o (PARI)

%o { for (n=2,1000, N = 2^n;

%o p1 = precprime(N-1);

%o p2 = nextprime(N+1);

%o ba = bitand(p1, p2);

%o if ( bitand( ba, ba-1 ) == 0, print1(n,", "));

%o ); }

%o /* _Joerg Arndt_, Aug 16 2012 */

%Y Cf. A007053, A175330.

%K nonn,base

%O 0,1

%A _Alex Ratushnyak_, Aug 07 2012

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Last modified January 24 16:34 EST 2020. Contains 331207 sequences. (Running on oeis4.)