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.

A364319 a(n) = (A077446(n) + 1)/2 for n >= 0.

Original entry on oeis.org

0, 1, 3, 6, 16, 33, 91, 190, 528, 1105, 3075, 6438, 17920, 37521, 104443, 218686, 608736, 1274593, 3547971, 7428870, 20679088, 43298625, 120526555, 252362878, 702480240, 1470878641, 4094354883, 8572908966, 23863649056, 49966575153, 139087539451
Offset: 0

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Author

Wolfdieter Lang, Aug 15 2023

Keywords

Comments

a(n) and b(n) = A006452(n+1), for n >= 0, give the nonnegative solution of the equation binomial(a(n), 2) = b(n)^2 - 1.
This shows that the number of independent elements of an antisymmetric a(n) X a(n) matrix coincides with the number of independent elements of a traceless b(n) X b(n) matrix. The n = 0 case is trivial: 0 = 0. (The question about this coincidence was posed to W. L. by Martin Bordemann, Mar 03 1991.)

Examples

			The solutions (a(n), b(n)) begin:
  n: 0 1 2 3  4  5  6   7   8    9   10   11    12    13     14     15 ...
  ------------------------------------------------------------------------
  a: 0 1 3 6 16 33 91 190 528 1105 3075 6438 17920 37521 104443 218686 ...
  b: 1 1 2 4 11 23 64 134 373  781 2174 4552 12671 26531  73852 154634 ...
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 3, 6, 16}, 31] (* Robert P. P. McKone, Aug 29 2023 *)

Formula

a(n) = 6*a(n-2) - a(n-4) - 2, for n >= 0, with a(-4) = -32, a(-3) = -15, a(-2) = -5, a(-1) = -2.
O.g.f.: G(x) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 6*x^2 + x^4)) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 2*x - x^2)*(1 + 2*x - x^2)).
Bisection: a(2*k) = (5*S(k-1, 6) + S(k-2, 6) + 1)/2 and a(2*k+1) = (S(k, 6) + 5*S(k-1, 6) + 1)/2, for k >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1).
Bisection: a(2*k) = (1 + 8*q(k) - p(k))/2 and a(2*k+1) = (1 + 8*q(k) + p(k))/2, for k >= 0, with p(k) = A001541(k) = S(k, 6) - 3*S(k-1, 6) and q(k) = A001109(k) = S(k-1, 6).
E.g.f.: (cosh(x) - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) + sinh(x) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x))/2. - Stefano Spezia, Aug 29 2023