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

1,2

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.

Formula

From Peter Bala, Dec 26 2023: (Start)

a(n) = A069097(2*n-1).

Conjecture: a(n) = (1/3)*Sum_{k = 1..4*n-2} (-1)^k*gcd(k,4*n-2)^2. (End)

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

Cf. A002117, A069097.

Sequence in context: A082108 A024846 A024842 * A031072 A193880 A138248

Adjacent sequences: A304272 A304273 A304274 * A304276 A304277 A304278

Keyword

nonn,easy

Author

Hugo Pfoertner, May 10 2018

Status

approved