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A060208
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a(n) = 2*pi(n) - pi(2*n), where pi(i) = A000720(i).
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10
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-1, 0, 1, 0, 2, 1, 2, 2, 1, 0, 2, 1, 3, 3, 2, 1, 3, 3, 4, 4, 3, 2, 4, 3, 3, 3, 2, 2, 4, 3, 4, 4, 4, 3, 3, 2, 3, 3, 3, 2, 4, 3, 5, 5, 4, 4, 6, 6, 5, 5, 4, 3, 5, 4, 3, 3, 2, 2, 4, 4, 6, 6, 6, 5, 5, 4, 6, 6, 5, 4, 6, 6, 8, 8, 7, 6, 6, 6, 7, 7, 7, 6, 8, 7, 7, 7, 6, 6, 8, 7, 6, 6, 6, 6, 6, 5, 6, 6, 5, 4, 6, 6, 8, 8, 8, 7, 9, 9, 11, 11, 11, 10, 12, 11, 10, 10, 9, 9, 9, 8, 7, 7
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OFFSET
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1,5
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COMMENTS
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Rosser & Schoenfeld show 2pi(x) > pi(2x) for x>10. - _N. J. A. Sloane_, Jul 03 2013, corrected Jul 09 2015
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REFERENCES
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J. Barkley Rosser and Lowell Schoenfeld, Abstracts of Scientific Communications, Internat. Congress Math., Moscow, 1966, Section 3, Theory of Numbers.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section VII.5, p. 235.
Sanford Segal, On Pi(x+y)<=Pi(x)+Pi(y). Transactions American Mathematical Society, 104 (1962), 523-527.
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LINKS
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Eugene Ehrhart, On prime numbers, Fibonacci Quarterly 26:3 (1988), pp. 271-274. Shows a(n)>0 for n>10.
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FORMULA
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a(n) = Mod[2*PrimePi[n], PrimePi[2n]] = 2*A000720(n)-A000720(2n) for n>1.
a(n) ~ 2n log 2 / (log n)^2, by the prime number theorem. - _N. J. A. Sloane_, Mar 12 2007
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EXAMPLE
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n=100, pi(100)=25, pi(200)=46, 2pi(100)-pi(2*100) =4=a(100)
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MATHEMATICA
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f[n_] := 2 PrimePi[n] - PrimePi[2 n]; Array[f, 122] (* _Robert G. Wilson v_, Aug 12 2011 *)
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PROG
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(PARI) a(n)=2*primepi(n)-primepi(2*n) \\ _Charles R Greathouse IV_, Jul 02 2013
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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_Labos Elemer_, Mar 19 2001
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EXTENSIONS
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Edited by _N. J. A. Sloane_, Jul 03 2013
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STATUS
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approved
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