OFFSET
1,6
COMMENTS
Sum_{k>0} mu(k)/k = limit_{n->oo} A070888(n)/A070889(n) = 0. - Jean-François Alcover, Apr 18 2013. This is equivalent to the Prime Number Theorem! - N. J. A. Sloane, Feb 04 2022
REFERENCES
Harold M. Edwards, Riemann's Zeta Function, Dover Publications, New York, 1974 (ISBN 978-0-486-41740-0), p. 92.
Edmund Landau, Handbuch der Lehre von der Verteilung der Primzahlen, Chelsea Publishing, NY 1953, p. 568.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..2376 (terms 1..1906 from Robert Israel)
F. K. Richter, A new elementary proof of the Prime Number Theorem, arXiv preprint arXiv:2002.03255 [math.NT], 2020-2021.
Harold N. Shapiro, Some assertions equivalent to the prime number theorem for arithmetic progressions, Communications on Pure and Applied Mathematics 2.2‐3 (1949): 293-308.
EXAMPLE
a(6) = 2 because 1-1/2-1/3-1/5+1/6 = 4/30 = 2/15.
MAPLE
T:= 0:
for n from 1 to 100 do
T:= T + numtheory:-mobius(n)/n;
A[n]:= numer(T)
od:
seq(A[n], n=1..100); # Robert Israel, Aug 04 2014
MATHEMATICA
Table[ Numerator[ Sum[ MoebiusMu[k]/k, {k, 1, n}]], {n, 1, 37}]
PROG
(PARI) t = 0; v = []; for( n = 1, 60, t= t + moebius( n) / n; v = concat( v, numerator( t))); v \\ adapted to latest PARI version by Michel Marcus, Aug 04 2014
(Python)
from functools import lru_cache
from sympy import harmonic
@lru_cache(maxsize=None)
def f(n):
if n <= 1:
return 1
c, j = 1, 2
k1 = n//j
while k1 > 1:
j2 = n//k1 + 1
c += (harmonic(j-1)-harmonic(j2-1))*f(k1)
j, k1 = j2, n//j2
return c+harmonic(j-1)-harmonic(n)
def A070888(n): return f(n).numerator # Chai Wah Wu, Nov 03 2023
CROSSREFS
KEYWORD
frac,sign
AUTHOR
Donald S. McDonald, May 17 2002
EXTENSIONS
Edited by Robert G. Wilson v, Jun 10 2002
STATUS
approved