%I #37 Jan 22 2024 06:02:35
%S 0,1,2,0,3,1,4,2,0,5,3,1,6,4,2,0,7,5,3,1,8,6,4,2,0,9,7,5,3,1,10,8,6,4,
%T 2,0,11,9,7,5,3,1,12,10,8,6,4,2,0,13,11,9,7,5,3,1,14,12,10,8,6,4,2,0,
%U 15,13,11,9,7,5,3,1,16,14,12,10,8,6,4,2,0,17,15,13,11,9,7,5,3,1,18,16,14
%N Irregular triangle read by rows: row n begins with n and decreases by 2 until 0 or 1 is reached, for n >= 0.
%C As a sequence, a(n) = A025644(n+1) for n <= 142.
%C The length of row n is given by A008619(n) = 1 + floor(n/2).
%C From _Wolfdieter Lang_, Feb 17 2020: (Start)
%C This table T(n, m) can be used for the conversion identity
%C 2*cos(Pi*k/N) = 2*sin((Pi/(2*N))*(N - 2*k)) = 2*sin((Pi/(2*N))*T(N-2, k-1)), here for N = n+2 >= 2, and k = m + 1 = 1, 2, ..., floor(N/2).
%C 2*cos((Pi/N)*k) = R(k, rho(N)), where R is a monic Chebyshev polynomial from A127672 and rho(N) = 2*cos(Pi/N), gives part of the roots of the polynomial S(N-1, x), for k = 1, 2, ..., floor(N/2), with the Chebyshev S polynomials from A049310.
%C 2*sin((Pi/(2*N))*q) = d^{(2*N)}_q/r, for q = 1, 2, ..., N, with the length ratio (q-th diagonal)/r, where r is the radius of the circle circumscribing a regular (2*N)-gon. The counting q starts with the diagonal d^{(2*N)}_1 = s(2*N) (in units of r), the side of the (2*N)-gon. The next diagonal is d^{(2*N)}_2 = rho(2*N)*s(2*N) (in units of r).
%C For the instances N = 4 (n = 2) and 5 (n = 3) see the example section. (End)
%C If b > 0 and c > 0 are the integer coefficients of a monic quadratic x^2 + b*x + c, it has integer roots if its discriminant d^2 = b^2 - 4c is a perfect square. This sequence is the values of d for increasing b sorted by b then c. The first pair of (b, c) = (2, 1) and has d = a(0) = 0. The n-th pair of (b, c) = (A027434(n), A350634(n)) and has d = a(n-1). - _Frank M Jackson_, Jan 20 2024
%H Harvey P. Dale, <a href="/A082375/b082375.txt">Table of n, a(n) for n = 0..1000</a>
%F T(n, m) = n - 2*m, m = 0, 1, ..., floor(n/2), n >= 0 (see the name and programs). - _Wolfdieter Lang_, Feb 17 2020
%F a(n) = A199474(n+1) - A122197(n+1). - _Wesley Ivan Hurt_, Jan 09 2022
%F a(n) = sqrt((A027434(n+1))^2 - 4*A350634(n+1)). - _Frank M Jackson_, Jan 20 2024
%e The irregular triangle T(n, m) begins:
%e n\m 0 1 2 3 4 5 ...
%e 0: 0
%e 1: 1
%e 2: 2 0
%e 3: 3 1
%e 4: 4 2 0
%e 5: 5 3 1
%e 6: 6 4 2 0
%e 7: 7 5 3 1
%e 8: 8 6 4 2 0
%e 9: 9 7 5 3 1
%e 10: 10 8 6 4 2 0
%e ... reformatted by _Wolfdieter Lang_, Feb 15 2020
%e From _Wolfdieter Lang_, Feb 17 2020: (Start)
%e Conversion identity:
%e N = n+2 = 4: k = m+1 = 1, 2*cos(Pi*1/4) = 2*sin(Pi*2/8) = sqrt(2); k=2, 2*cos(Pi*2/4) = 2*sin(Pi*0/8) = 0.
%e N = 5:(n=3) k=1 (m=0), 2*cos(Pi*1/5) = 2*sin(Pi*3/10) = (1 + sqrt(5))/2 = rho(5) = A001622; k=2: 2*cos(Pi*2/5) = 2*sin(Pi*1/10) = rho(5) - 1. (End)
%t Flatten[Table[Range[n,0,-2],{n,0,20}]] (* _Harvey P. Dale_, Apr 03 2019 *)
%t lst = {}; Do[If[IntegerQ[d=Sqrt[b^2-4c]], AppendTo[lst, d]], {b, 1, 20}, {c, 1, b^2/4}]; lst (* _Frank M Jackson_, Jan 20 2024 *)
%o (PARI) a(n)=local(m); if(n<0,0,m=sqrtint(1+4*n); m-1-(1+4*n-m^2)\2)
%Y Cf. A025644, A008619, A008619, A027434, A122197, A199474, A350634.
%K nonn,tabf
%O 0,3
%A _Michael Somos_, Apr 09 2003