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%I #32 Oct 16 2023 12:41:19
%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 */
%o (Python)
%o from itertools import islice
%o from sympy import prevprime, nextprime
%o def A214415_gen(): # generator of terms
%o n, m = 2, 4
%o while True:
%o if prevprime(m)&nextprime(m) == 1:
%o yield n
%o n += 1
%o m *= 2
%o A214415_list = list(islice(A214415_gen(),20)) # _Chai Wah Wu_, Oct 16 2023
%Y Cf. A007053, A175330.
%K nonn,base
%O 0,1
%A _Alex Ratushnyak_, Aug 07 2012
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