A384717 - OEIS

A384717

Sum of floored squared chord lengths from -1 to the 2n-th roots of unity (upper semicircle, endpoints excluded).

0, 2, 4, 5, 6, 9, 9, 11, 13, 14, 15, 18, 18, 20, 22, 23, 24, 27, 27, 29, 31, 32, 33, 36, 36, 38, 40, 41, 42, 45, 45, 47, 49, 50, 51, 54, 54, 56, 58, 59, 60, 63, 63, 65, 67, 68, 69, 72, 72, 74, 76, 77, 78, 81, 81, 83, 85, 86, 87, 90, 90, 92, 94, 95, 96, 99, 99

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OFFSET

1,2

COMMENTS

Let z_k = exp(i*Pi*k/n) for k = 1..n-1 (upper 2n-th roots, excluding endpoints +-1). By Euler's formula, |1 + z_k|^2 = 2 + 2*cos(Pi*k/n). The sequence a(n) is Sum_{k=1..n-1} floor(2 + 2*cos(Pi*k/n)), i.e., the sum of floored squared chord lengths from -1 to those points.

Distribution of the summands: floor(2 + 2*cos(Pi*k/n)) takes values 3, 2, 1, 0 exactly floor(n/3), floor(n/2) - floor(n/3), floor(2*n/3) - floor(n/2), (n-1) - floor(2*n/3) times, respectively.

If one includes the endpoints (k = 0 and k = n), the resulting total is a(n) + 4.

LINKS

Paolo Xausa, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (0,1,1,0,-1).

FORMULA

a(n) = floor(n/2) + floor(n/3) + floor(2*n/3).

From Elmo R. Oliveira, Apr 12 2026: (Start)

a(n) = a(n-2) + a(n-3) - a(n-5).

G.f.: x^2*(2 + 4*x + 3*x^2)/((1 - x^2)*(1 - x^3)). (End)

EXAMPLE

n = 10: the list floor(2 + 2*cos(Pi*k/10)) for k = 1..9 is 3, 3, 2, 2, 2, 1, 1, 0, 0; sum = 14.

Check with the closed form: floor(10/2) + floor(10/3) + floor(20/3) = 5 + 3 + 6 = 14.