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A259922
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a(n)= Sum_{2 < prime p <= n} c_p - Sum_{n < prime p < 2*n} c_p, where 2^c_p is the greatest power of 2 dividing p-1.
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1
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0, -1, -1, -2, 2, 1, 1, 1, -3, -4, -2, -3, 1, 1, -1, -2, 6, 6, 6, 6, 3, 2, 4, 3, 3, 3, 1, 1, 5, 4, 4, 4, 4, 3, 3, 2, 3, 3, 3, 2, 8, 7, 9, 9, 6, 6, 8, 8, 3, 3, 1, 0, 4, 3, 1, 1, -3, -3, -1, -1, 3, 3, 3, 2, 2, 1, 3, 3, 0, -1, 1, 1, 7, 7, 5, 4, 4, 4, 4, 4, 4, 3
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OFFSET
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1,4
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COMMENTS
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It is known that, for n>10, pi(2*n) < 2*pi(n), where pi(n) is the number of primes not exceeding n (A000720). Thus, for n>10, in the interval (1,n] we have more primes than in the interval (n,2*n).
In connection with this, it is natural to conjecture that there exists a number N such that a(n)>0 for all n >= N.
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LINKS
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MATHEMATICA
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Map[Total[Flatten[Map[IntegerExponent[Select[#, PrimeQ]-1, 2]&, {Range[3, #], Range[#+1, 2#-1]}]{1, -1}]]&, Range[50]]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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