

A305445


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


2



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
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OFFSET

2,7


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


LINKS



EXAMPLE

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).)


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:


PROG

(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))))}


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



