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

1,4

Comments

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

Links

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

Example

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

Prog

(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

Crossrefs

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

Sequence in context: A066416 A292342 A091991 * A324572 A153148 A378437

Adjacent sequences: A108231 A108232 A108233 * A108235 A108236 A108237

Keyword

easy,nonn,changed

Author

Mike Stay, Jun 16 2005

Extensions

Definition clarified by Antti Karttunen, Dec 16 2017

Status

approved