A068340 a(n) = Sum_{k=1..n} mu(k)*k, where mu(k) is the Moebius function.
1, -1, -4, -4, -9, -3, -10, -10, -10, 0, -11, -11, -24, -10, 5, 5, -12, -12, -31, -31, -10, 12, -11, -11, -11, 15, 15, 15, -14, -44, -75, -75, -42, -8, 27, 27, -10, 28, 67, 67, 26, -16, -59, -59, -59, -13, -60, -60, -60, -60, -9, -9, -62, -62, -7, -7, 50, 108, 49
Offset: 1
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Haskell
a068340 n = a068340_list !! (n-1) a068340_list = scanl1 (+) a055615_list -- Reinhard Zumkeller, Sep 04 2015
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Maple
with(numtheory): a:= proc(n) a(n):= mobius(n)*n +a(n-1) end: a(0):=0: seq(a(n), n=1..100); # Alois P. Heinz, Oct 21 2012
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Mathematica
Table[Sum[MoebiusMu[k]k,{k,n}],{n,60}] (* Harvey P. Dale, Feb 01 2012 *)
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PARI
a(n) = sum(k=1, n, k*moebius(k)); \\ Michel Marcus, Jan 14 2023
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A068340(n): if n <= 1: return 1 c, j = 1, 2 k1 = n//j while k1 > 1: j2 = n//k1 + 1 c -= (j2*(j2-1)-j*(j-1)>>1)*A068340(k1) j, k1 = j2, n//j2 return c-(n*(n+1)-(j-1)*j>>1) # Chai Wah Wu, Apr 04 2023
Formula
G.f. A(x) satisfies x = Sum_{k>=1} k * (1 - x^k) * A(x^k). - Seiichi Manyama, Apr 01 2023
Sum_{k=1..n} k * a(floor(n/k)) = 1. - Seiichi Manyama, Apr 03 2023
Comments