A058290 Rounded difference between 10^n/(log(10^n) - A) and pi(10^n), where A is Legendre's constant and pi is the prime counting function.
-1, 4, 3, 4, 2, -4, 45, 561, 6549, 69985, 690493, 6545056, 60615397, 555560046, 5070271362, 46223804313, 421692578206, 3853431791690, 35289854434775, 323979090116197, 2981921009910364, 27516571651291205, 254562416350667928
Offset: 0
References
- Jan Gullberg, "Mathematics, From the Birth of Numbers", W. W. Norton and Company, NY and London, 1997, page 81.
Links
- Eduard Roure Perdices, Table of n, a(n) for n = 0..28 (terms 0..23 from Harry J. Smith).
Programs
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Mathematica
Table[ Round[ 10^n /(Log[10^n] - 1.08366) - PrimePi[10^n] ], {n, 0, 13} ]
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PARI
{A006880_vec = [0, 4, 25, 168, 1229, 9592, 78498, 664579, 5761455, 50847534, 455052511 4118054813, 37607912018, 346065536839, 3204941750802, 29844570422669, 279238341033925, 2623557157654233, 24739954287740860, 234057667276344607, 2220819602560918840, 21127269486018731928, 201467286689315906290, 1925320391606803968923]} \\ Edited by M. F. Hasler, Dec 03 2018 {default(realprecision, 100); t=log(10); for (n=0, 23, write("b058290.txt", n, " ", round(10^n/(n*t - 1.08366)) - A006880_vec[n+1]))} \\ Harry J. Smith, Jun 22 2009
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PARI
A058290(n)={10^n\/(n*log(10)-1.08366)-A006880(n)} \\ with A006880(n)=primepi(10^n) and/or precomputed values for n > 10. - M. F. Hasler, Dec 03 2018
Formula
a(n) = round(10^n/(log(10^n) - 1.08366)) - A006880(n). - M. F. Hasler, Dec 03 2018
Extensions
More terms from Harry J. Smith, Jun 22 2009
Comments