OFFSET
1,4
COMMENTS
FORMULA
a(n) = d(floor((n+1)/2))^((n+1) mod 2), for n >= 2.
a(n) = d( (n+2+(n-2)*(-1)^n)/4 ) for n >= 2.
a(n) = Sum_{k=1..floor(n/2)} c(n/k) * c(floor(n/2)/k), where c(m) = 1 - ceiling(m) + floor(m).
Sum_{k=1..n} a(k) ~ (log(n/2) + 2*gamma)*n/2, where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 28 2025
MAPLE
with(numtheory): 0, seq(2*tau(n) - tau(2*n) + (n mod 2), n=2..100); # Ridouane Oudra, Jan 18 2025
MATHEMATICA
Join[{0}, Table[DivisorSigma[0, (n+2+(n-2)*(-1)^n)/4], {n, 2, 100}]]
PROG
(PARI) a(n) = if(n == 1, 0, numdiv((n+2+(n-2)*(-1)^n)/4)); \\ Amiram Eldar, Jan 28 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Jan 03 2024
STATUS
approved