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%I #21 Dec 28 2023 01:52:45
%S 1,11,29,55,105,131,181,319,305,379,605,551,745,963,869,991,1441,1595,
%T 1405,1991,1721,1891,3045,2255,2737,3355,2861,3799,4169,3539,3781,
%U 5775,5249,4555,6061,5111,5401,8195,7205,6319,8721,6971,8845
%N a(n) = Sum_{k = 1..n} gcd(k,n) / cos(Pi*k/n)^2 for odd n.
%H Hugo Pfoertner, <a href="/A304275/b304275.txt">Table of n, a(n) for n = 1..1000</a>
%H Nikolai Osipov (Proposer), <a href="http://dx.doi.org/10.4169/amer.math.monthly.124.8.754">Problem 12003</a>, Amer. Math. Monthly 124 (No. 8, Oct. 2017), page 754.
%F From _Peter Bala_, Dec 26 2023: (Start)
%F a(n) = A069097(2*n-1).
%F Conjecture: a(n) = (1/3)*Sum_{k = 1..4*n-2} (-1)^k*gcd(k,4*n-2)^2. (End)
%F Sum_{k=1..n} a(k) ~ c * n^3, where c = 4*Pi^2 / (21*zeta(3)) = 1.563923... . - _Amiram Eldar_, Dec 28 2023
%p 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
%t 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 *)
%o (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
%Y Cf. A002117, A069097.
%K nonn,easy
%O 1,2
%A _Hugo Pfoertner_, May 10 2018
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