OFFSET
0,1
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 = 3+sqrt(8), and the limit p3(r) = 5.854315472394508538153482993682502287049948...
REFERENCES
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.~
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
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=3+sqrt(8) and a(0) = ceiling(x).
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3).
G.f.: (6 + 4*x - x^2)/(1 - 5*x - 5*x^2 + x^3).
a(n) = (1/16)*(2*(-1)^n + (47-33*sqrt(2))*(3-2*sqrt(2))^n + (3+2*sqrt(2))^n*(47+33*sqrt(2))). - Colin Barker, Nov 13 2017
EXAMPLE
a(0) = ceiling(r) = 6, where r = 3+sqrt(8);
a(1) = floor(6*r) = 34; a(2) = ceiling(35*r) = 199.
MATHEMATICA
x = 3 + Sqrt[8]; z = 30; (* z = # terms in sequences *)
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]]
t1 = Table[p1[n], {n, 0, z}] (* A001653 *)
t2 = Table[p2[n], {n, 0, z}] (* A084158 *)
t3 = Table[p3[n], {n, 0, z}] (* A218990 *)
t4 = Table[p4[n], {n, 0, z}] (* A001109 *)
LinearRecurrence[{5, 5, -1}, {6, 34, 199}, 30] (* Harvey P. Dale, Mar 21 2024 *)
PROG
(PARI) Vec((6 + 4*x - x^2) / ((1 + x)*(1 - 6*x + x^2)) + O(x^50)) \\ Colin Barker, Nov 13 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 11 2012
STATUS
approved