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.

A214998 Power ceiling-floor sequence of 2 + sqrt(3).

Original entry on oeis.org

4, 14, 53, 197, 736, 2746, 10249, 38249, 142748, 532742, 1988221, 7420141, 27692344, 103349234, 385704593, 1439469137, 5372171956, 20049218686, 74824702789, 279249592469, 1042173667088, 3889445075882, 14515606636441, 54172981469881, 202176319243084
Offset: 0

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Author

Clark Kimberling, Nov 10 2012

Keywords

Comments

See A214992 for a discussion of power ceiling-floor sequence and power ceiling-floor function, p3(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = 2 + sqrt(3), and the limit p3(r) = (23 + 13*sqrt(3))/12.

Examples

			a(0) = ceiling(r) =  4, where r = 2+sqrt(3);
a(1) = floor(4*r) = 14; a(2) = ceiling(14*r) = 53.
		

Crossrefs

Programs

  • Mathematica
    x = 2 + Sqrt[3]; z = 30; (* z = # terms in sequences *)
    z1 = 100; (* z1 = # digits in approximations *)
    f[x_] := Floor[x]; c[x_] := Ceiling[x];
    p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
    p1[n_] := f[x*p1[n - 1]]
    p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
    p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
    p4[n_] := c[x*p4[n - 1]]
    Table[p1[n], {n, 0, z}]  (* A001835 *)
    Table[p2[n], {n, 0, z}]  (* A109437 *)
    Table[p3[n], {n, 0, z}]  (* A214998 *)
    Table[p4[n], {n, 0, z}]  (* A001353 *)
  • PARI
    Vec((4 + 2*x - x^2) / ((1 + x)*(1 - 4*x + x^2)) + O(x^30)) \\ Colin Barker, Nov 13 2017

Formula

a(n) = floor(x*a(n-1)) if n is odd, a(n) = ceiling(x*a(n-1)) if n is even, where x = 2+sqrt(3) and a(0) = ceiling(x).
a(n) = 3*a(n-1) + 3*a(n-2) - a(n-3).
G.f.: (4 + 2*x - x^2)/(1 - 3*x - 3*x^2 + x^3).
a(n) = (-1)^n + 4*a(n-1) - a(n-2) with a(0) = 4 and a(1) = 14. - Peter Bala, Nov 12 2017
a(n) = (1/12)*(2*(-1)^n + (23-13*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(23+13*sqrt(3))). - Colin Barker, Nov 13 2017