A305445 Minimum number of bit inversions to convert n into a prime.

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

Offset: 2

Author

Rick L. Shepherd, Aug 12 2018

Keywords

• nonn
• base

Comments

If n is already a prime, a(n) is defined to be 0. Every original bit of n's binary representation is allowed to be inverted, but no leading 0 bits may be. Every n > 1 is either a prime or can be converted to a prime by bit inversions (guaranteed because, say, 0...010 is the prime 2). The maximum value of the first 10^7 terms is 3.

This sequence was inspired by the linked "code golf" problem, which converts n to a square but (unlike this sequence) disallows inverting n's most significant bit.

The least n for which a(n) = 4 is n = 45812984490. - Giovanni Resta, Jan 03 2019

Examples

			For n = 8, the binary representation 1000 cannot be turned into a prime with only one bit inversion, but 0010, where both the first and third bits from the left are inverted, is the prime 2, so a(8) = 2. (There are other primes possible with two inversions in this case: 1011 (11 decimal) and 1101 (13 decimal).)
		

Links

• Robert Israel, Table of n, a(n) for n = 2..10000
• Programing Puzzles & Code Golf Stack Exchange, Toggle some bits and get a square

Programs

Maple

f:= proc(n) local m,d,x;
  if isprime(n) then return 0 fi;
  m:= ilog2(n);
  for d from 1 do
    for x in combinat:-choose([$0..m],d) do
      if isprime(Bits:-Xor(n, add(2^i,i=x))) then return d fi
  od od
end proc:
map(f, [$2..200]); # Robert Israel, Aug 20 2018

PARI

{a(n) = my(b, L, N, s, v); if(n < 2, ,
if(isprime(n), 0, b = binary(n); L = #b; for(j = 1, L, v = vector(j, Y, [1, L]);
forvec(X = v, N = n + sum(k = 1, j, if(b[X[k]], s = -1, s = 1); s*2^(L - X[k])); if(isprime(N), return(j)), 2))))}