OFFSET
1,2
LINKS
Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Nikolai Osipov (Proposer), Problem 12003, Amer. Math. Monthly 124 (No. 8, Oct. 2017), page 754.
Nikolay Osipov, Proposer's solution to Problem 12003
FORMULA
a(n) = A069097(2*n-1). - Peter Bala, Dec 26 2023
a(n) = (1/3)*Sum_{k = 1..4*n-2} (-1)^k*gcd(k,4*n-2)^2. - Conjectured by Peter Bala, Dec 26 2023; proved by Nikolay Osipov, Oct 05 2024
Sum_{k=1..n} a(k) ~ c * n^3, where c = 4*Pi^2 / (21*zeta(3)) = 1.563923... . - Amiram Eldar, Dec 28 2023
MAPLE
seq( round( add(igcd(k, 2*n+1)/cos(Pi*k/(2*n+1))^2, k = 1..2*n+1) ), n = 0..40); # Peter Bala, Dec 26 2023
MATHEMATICA
f[p_, e_] := p^(e-1)*(p^e*(p+1)-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[2*n - 1]; Array[a, 50] (* Amiram Eldar, Dec 28 2023 *)
PROG
(PARI) a(n) = n = 2*n-1; round(sum(k=1, n, gcd(k, n) / cos(Pi*k/n)^2)); \\ Michel Marcus, May 10 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Hugo Pfoertner, May 10 2018
STATUS
approved