OFFSET
1,3
COMMENTS
a(n) = number of terms among {ceiling(n/k)}, 1 <= k <= n, that are odd.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000
FORMULA
MAPLE
b:= n-> add((-1)^d, d=numtheory[divisors](n)):
a:= proc(n) option remember; `if`(n>0, 1+b(n-1)+a(n-1), 0) end:
seq(a(n), n=1..80); # Alois P. Heinz, May 25 2020
MATHEMATICA
Table[Sum[Mod[Ceiling[n/k], 2], {k, 1, n}], {n, 1, 80}]
Table[n - Sum[DivisorSum[k, (-1)^(# + 1) &], {k, 1, n - 1}], {n, 1, 80}]
nmax = 80; CoefficientList[Series[x/(1 - x) (1 + Sum[x^(2 k)/(1 + x^k), {k, 1, nmax}]), {x, 0, nmax}], x] // Rest
PROG
(PARI) a(n) = sum(k=1, n, ceil(n/k) % 2); \\ Michel Marcus, May 25 2020
(Python)
from math import isqrt
def A330926(n): return n+(s:=isqrt(n-1))**2-((t:=isqrt(m:=n-1>>1))**2<<1)-(sum((n-1)//k for k in range(1, s+1))-(sum(m//k for k in range(1, t+1))<<1)<<1) # Chai Wah Wu, Oct 23 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, May 25 2020
STATUS
approved