A214993 Power floor sequence of (golden ratio)^5.

11, 121, 1341, 14871, 164921, 1829001, 20283931, 224952241, 2494758581, 27667296631, 306835021521, 3402852533361, 37738212888491, 418523194306761, 4641493350262861, 51474950047198231, 570865943869443401, 6331000332611075641, 70211869602591275451

Offset: 0

Author

Clark Kimberling, Nov 09 2012

Keywords

• nonn
• easy

Comments

See A214992 for a discussion of power floor sequence and also the power floor function, p1(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = (golden ratio)^5, and the limit p1(r) = (3/22)*(3+2*sqrt(5)).

Examples

			a(0) = [r] = [11.0902] = 11, where r = (1+sqrt(5))^5.
a(1) = [11*r] = 121; a(2) = [121*r] = 1341.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..250
• Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
• Index entries for linear recurrences with constant coefficients, signature (12,-10,-1).

Crossrefs

Cf. A214992, A214994, A049666, A015457.

Programs

Magma

I:=[11,121,1341]; [n le 3 select I[n] else 12*Self(n-1)-10*Self(n-2)-Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 01 2018

Mathematica

x = GoldenRatio^5; 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}]  (* A214993 *)
Table[p2[n], {n, 0, z}]  (* A049666 *)
Table[p3[n], {n, 0, z}]  (* A015457 *)
Table[p4[n], {n, 0, z}]  (* A214994 *)
LinearRecurrence[{12,-10,-1}, {11,121,1341}, 30] (* G. C. Greubel, Feb 01 2018 *)

PARI

Vec((11 - 11*x - x^2) / ((1 - x)*(1 - 11*x - x^2)) + O(x^20)) \\ Colin Barker, Nov 13 2017

Formula

a(n) = [x*a(n-1)], where x=((1+sqrt(5))/2)^5, a(0) = [x].

a(n) = 1 (mod 10).

a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3).

G.f.: (11 - 11*x - x^2)/(1 - 12*x + 10*x^2 + x^3).

a(n) = (1/55)*(5 + (300-134*sqrt(5))*((11-5*sqrt(5))/2)^n + 2*(11/2+(5*sqrt(5))/2)^n*(150+67*sqrt(5))). - Colin Barker, Nov 13 2017