OFFSET
1,7
COMMENTS
The length of row n is A232626(n).
In a regular n-gon, n>=2, inscribed in a circle of radius R (in some length units), 2*sin(4*Pi/n) = (S(n)/R)*(D(1,n)/S(n)) = D(1,n)/R, with the side length S(n) and the length of the first (smallest) diagonal D(1,n). For n=2 there is no such diagonal, and one can put D(1,2) = 0. Obviously, D(1,2*m) = S(m), m >= 2.
See a comment on A231190 regarding the pair of sequences p(k,n) and q(k,n), n >= 1, k >= 1. Here k=2 with A231190 and A232625.
See also the k=1 analog A231189 of the present table.
The relevant identity is here 2*sin(Pi*4/n) = 2*cos(Pi*abs(n-8)/(2*n)) = 2*cos(Pi*p(2,n)/q(2,n)). This is R(p(2,n), x) (mod C(q(2,n), x)), with x = 2*cos(Pi/q(2,n)) =: rho(q(2,n)). with the coefficient tables for the polynomials R and C given in A127672 and A187360, respectively. This gives the power base coefficients of 2*sin(Pi*4/n) in the algebraic number field Q(rho(q(2,n))) of degree delta(q(2,n)), with delta(n) = A055034(n), shown in A232626.
If the degree p(2,n) of R(p(2,n), x) is smaller than the degree A232626(n) of C(q(2,n), x) then 2*sin(Pi*4/n) = R(p(2,n), x). Otherwise the (mod C(q(2,n), x)) congruence is needed. This happens for n = 1, 2, 3, 4, 21, 24, 27, 30,...
The power basis of Q(rho(q(2,n))) is <1, rho(q(2,n)), ..., rho(q(2,n))^(delta(q(2,n))-1)>. Therefore the length of row n of this table is delta(q(2,n)) = A232626(n).
The coefficient table for the minimal polynomial of 2*sin(Pi*4/n) is given in A232630.
FORMULA
EXAMPLE
The table a(n,m) begins (the trailing zeros are needed to have the correct degree A232626(n) in Q(rho(q(2,n))))
-----------------------------------------------------------------
n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ...
1: 0
2: 0
3: 0 -1
4: 0
5: 0 -3 0 1
6: 0 1
7: 0 1 0 0 0 0
8: 2
9: 0 1 0 0 0 0
10: 0 1 0 0
11: 0 -3 0 1 0 0 0 0 0 0
12: 0 1
13: 0 5 0 -5 0 1 0 0 0 0 0 0
14: 0 -3 0 1 0 0
15: 0 -7 0 14 0 -7 0 1
16: 0 1
17: 0 9 0 -30 0 27 0 -9 0 1 0 0 0 0 0 0
18: 0 5 0 -5 0 1
19: 0 -11 0 55 0 -77 0 44 0 -11 0 1 0 0 0 0 0 0
20: 0 -3 0 1
...
n=1: 2*sin(Pi*4/1) = 0. R(p(2,1), x) = R(7, x) = -7*x + 14*x^3 -7*x^5 + x^7. C(q(2,1), x) = C(2, x) = x, hence R(7, x) (mod C(2, x)) == 0, and
with A232626(1) = 1, a(1,0) = 0.n=7: p(2,7) = A231190(7) = 1, q(2,7) = A232625(7) = 14. 2*sin(Pi*4/7) = R(1, x) = x = rho(14) := 2*cos(Pi/14). C(14, x) of degree 6 does not apply here. A232626(7) = 6, hence the row n=7 is 0 1 0 0 0 0.
n=9: p(2,9) = 1, q(2,9) = 18. 2*sin(Pi*4/9) = R(1, x) = x = rho(18) = 2*cos(Pi/18). C(18, x) with degree 6 is not needed here. A232626(9) = 6, hence row n=9 is also 0 1 0 0 0 0.
n=8: this row with entry 2 coincides with row n=4 of A231189.
n=17: row length A232626(17) = 16; p(2,17) = 9; C(34, x) has degree 16, therefore the R(9, x) coefficients produce here the first 10 entries for row n=17: 0 9 0 -30 0 27 0 -9 0 1, followed by 6 zeros, and 2*sin(Pi*4/17) = 9*rho(34) - 30*rho(34)^3 + 27*rho(34)^5 - 9*rho(34)^7 + 1*rho(34)^9, with rho(34) = 2*cos(Pi/34).
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Wolfdieter Lang, Dec 17 2013
STATUS
approved