|
|
A068340
|
|
a(n) = Sum_{k=1..n} mu(k)*k, where mu(k) is the Moebius function.
|
|
17
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
|
|
LINKS
|
|
|
FORMULA
|
G.f. A(x) satisfies x = Sum_{k>=1} k * (1 - x^k) * A(x^k). - Seiichi Manyama, Apr 01 2023
|
|
MAPLE
|
with(numtheory):
a:= proc(n) a(n):= mobius(n)*n +a(n-1) end: a(0):=0:
|
|
MATHEMATICA
|
Table[Sum[MoebiusMu[k]k, {k, n}], {n, 60}] (* Harvey P. Dale, Feb 01 2012 *)
|
|
PROG
|
(Haskell)
a068340 n = a068340_list !! (n-1)
a068340_list = scanl1 (+) a055615_list
(PARI) a(n) = sum(k=1, n, k*moebius(k)); \\ Michel Marcus, Jan 14 2023
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|