|
|
A347030
|
|
a(n) = 1 + Sum_{k=2..n} (-1)^k * a(floor(n/k)).
|
|
4
|
|
|
1, 2, 1, 3, 2, 1, 0, 4, 4, 3, 2, 0, -1, -2, -1, 7, 6, 6, 5, 3, 4, 3, 2, -2, -2, -3, -3, -5, -6, -5, -6, 10, 11, 10, 11, 11, 10, 9, 10, 6, 5, 6, 5, 3, 3, 2, 1, -7, -7, -7, -6, -8, -9, -9, -8, -12, -11, -12, -13, -11, -12, -13, -13, 19, 20, 21, 20, 18, 19, 20, 19, 19, 18, 17, 17
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
|
|
LINKS
|
|
|
FORMULA
|
G.f. A(x) satisfies: A(x) = (1/(1 - x)) * (x + Sum_{k>=2} (-1)^k * (1 - x^k) * A(x^k)).
|
|
MATHEMATICA
|
a[n_] := a[n] = 1 + Sum[(-1)^k a[Floor[n/k]], {k, 2, n}]; Table[a[n], {n, 1, 75}]
nmax = 75; A[_] = 0; Do[A[x_] = (1/(1 - x)) (x + Sum[(-1)^k (1 - x^k) A[x^k], {k, 2, nmax}]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] // Rest
|
|
PROG
|
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
if n <= 1:
return n
c, j = 1, 2
k1 = n//j
while k1 > 1:
j2 = n//k1 + 1
c += (j2-j&1)*(-1 if j&1 else 1)*A347030(k1)
j, k1 = j2, n//j2
return c+(n+1-j&1)*(-1 if j&1 else 1) # Chai Wah Wu, Apr 04 2023
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|