A108234 Minimum m such that n*2^m+k is prime, for k < 2^m. In other words, assuming you've read n out of a binary stream, a(n) is the minimum number of additional bits (appended to the least significant end of n) you must read before it is possible to obtain a prime.

1, 0, 0, 2, 0, 1, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 2, 2, 1, 2, 2, 0, 1, 0, 2, 1, 2, 1, 1, 0, 3, 1, 2, 0, 3, 0, 1, 2, 3, 0, 1, 2, 1, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 0, 2, 1, 2, 1, 4, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 2, 1, 0, 3, 1, 2, 0, 2, 3, 1, 2, 2, 0, 1, 2, 3, 2, 2, 1, 1, 0, 1, 1, 2

Offset: 1

Author

Mike Stay, Jun 16 2005

Keywords

• easy
• nonn

Comments

Somewhat related to the Riesel problem, A040081, the minimum m such that n*2^m-1 is prime.

Examples

			a(12) = 3 because 12 = 1100 in binary and 97 = 1100001 is the first prime that starts with 1100, needing 3 extra bits.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A040081, A091991, A164022 (smallest prime).

Programs

MATLAB

% and Octave.
for n=1:100;m=0;k=0;while(~isprime(n*2^m+k))k=k+1;if k==2^m k=0;m=m+1;end;end;x(n)=m;end;x

PARI

A108234(n) = { my(m=0,k=0); while(!isprime((n*2^m)+k), k=k+1; if(2^m==k, k=0; m=m+1)); m; }; \\ Antti Karttunen, Dec 16 2017, after Octave/MATLAB code

Extensions

Definition clarified by Antti Karttunen, Dec 16 2017