cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A082375 Irregular triangle read by rows: row n begins with n and decreases by 2 until 0 or 1 is reached, for n >= 0.

Original entry on oeis.org

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, 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, 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
Offset: 0

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Author

Michael Somos, Apr 09 2003

Keywords

Comments

As a sequence, a(n) = A025644(n+1) for n <= 142.
The length of row n is given by A008619(n) = 1 + floor(n/2).
From Wolfdieter Lang, Feb 17 2020: (Start)
This table T(n, m) can be used for the conversion identity
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).
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.
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).
For the instances N = 4 (n = 2) and 5 (n = 3), we have:
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.
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)
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
This sequence is related to an instance of Clark Kimberling's generic dispersion arrays; in this case the leader sequence is the square numbers A000290 (without 0), and the follower sequence is the nonsquare numbers A000037. This sequence gives the 0-origin column index of n in the resulting dispersion array. - Allan C. Wechsler, Feb 26 2025

Examples

			The irregular triangle T(n, m) begins:
  n\m  0 1 2 3 4 5 ...
  0:   0
  1:   1
  2:   2 0
  3:   3 1
  4:   4 2 0
  5:   5 3 1
  6:   6 4 2 0
  7:   7 5 3 1
  8:   8 6 4 2 0
  9:   9 7 5 3 1
  10: 10 8 6 4 2 0
  ...
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[Range[n,0,-2],{n,0,20}]] (* Harvey P. Dale, Apr 03 2019 *)
    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 *)
  • PARI
    a(n)=local(m); if(n<0,0,m=sqrtint(1+4*n); m-1-(1+4*n-m^2)\2)

Formula

T(n, m) = n - 2*m, m = 0, 1, ..., floor(n/2), n >= 0 (see the name and programs). - Wolfdieter Lang, Feb 17 2020
a(n) = A199474(n+1) - A122197(n+1). - Wesley Ivan Hurt, Jan 09 2022
a(n) = sqrt((A027434(n+1))^2 - 4*A350634(n+1)). - Frank M Jackson, Jan 20 2024