A304275 a(n) = Sum_{k = 1..n} gcd(k,n) / cos(Pi*k/n)^2 for odd n.

1, 11, 29, 55, 105, 131, 181, 319, 305, 379, 605, 551, 745, 963, 869, 991, 1441, 1595, 1405, 1991, 1721, 1891, 3045, 2255, 2737, 3355, 2861, 3799, 4169, 3539, 3781, 5775, 5249, 4555, 6061, 5111, 5401, 8195, 7205, 6319, 8721, 6971, 8845

Offset: 1

Author

Hugo Pfoertner, May 10 2018

Keywords

• nonn
• easy

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

Crossrefs

Cf. A002117, A069097.

Programs

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 *)

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

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