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A305445
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Minimum number of bit inversions to convert n into a prime.
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2
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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
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2,7
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COMMENTS
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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
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LINKS
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EXAMPLE
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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).)
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MAPLE
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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:
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PROG
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(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))))}
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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