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A130911
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a(n) is the number of primes with odd binary weight among the first n primes minus the number with an even binary weight.
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6
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1, 0, -1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 8, 9, 10, 11, 12, 13, 14, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 21, 20, 19, 20, 19, 18, 19, 18, 19, 18, 19, 18, 19, 18, 17, 16, 15, 14, 15, 14, 15, 14, 13, 14, 13, 14, 15, 16, 17, 18, 19, 20, 19, 20, 19, 20, 19, 18, 19, 20, 21, 20, 19
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OFFSET
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1,6
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COMMENTS
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Shevelev conjectures that a(n) >= 0 for n > 3. Surprisingly, the conjecture also appears to be true if we count zeros instead of ones in the binary representation of prime numbers.
The conjecture is true for primes up to at least 10^13. Mauduit and Rivat prove that half of all primes are evil. - T. D. Noe, Feb 09 2009
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LINKS
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FORMULA
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a(n) = (number of odious primes <= prime(n)) - (number of evil primes <= prime(n)).
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MATHEMATICA
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cnt=0; Table[p=Prime[n]; If[EvenQ[Count[IntegerDigits[p, 2], 1]], cnt--, cnt++ ]; cnt, {n, 10000}]
Accumulate[If[OddQ[DigitCount[#, 2, 1]], 1, -1]&/@Prime[Range[100]]] (* Harvey P. Dale, Aug 09 2013 *)
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PROG
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(PARI)f(p)={v=binary(p); s=0; for(k=1, #v, if(v[k]==1, s++)); return(s%2)}; nO=0; nE=0; forprime(p=2, 520, if(f(p), nO++, nE++); an=nO-nE; print1(an, ", ")) \\ Washington Bomfim, Jan 14 2011
(Python)
from sympy import nextprime
from itertools import islice
def agen():
p, evod = 2, [0, 1]
while True:
yield evod[1] - evod[0]
p = nextprime(p); evod[bin(p).count('1')%2] += 1
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CROSSREFS
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Cf. A156549 (race between primes having an odd/even number of zeros in binary).
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KEYWORD
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nice,sign,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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