OFFSET
0,1
COMMENTS
See A214992 for a discussion of power floor sequence and the power floor function, p1(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = 2+sqrt(8), and the limit p1(r) = 3.8983688904482395322594950087206...
See A218989 for the power floor function, p4. For comparison with p1, limit(p4(r)/p1(r) = 4/3.
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (5,0,-4).
FORMULA
a(n) = floor(x*a(n-1)), where x=2+sqrt(8), a(0) = floor(x).
a(n) = 5*a(n-1) - 4*a(n-3).
G.f.: (4 - x - 4*x^2) / ((1 - x)*(1 - 4*x - 4*x^2)). [Corrected by Colin Barker, Nov 13 2017]
a(n) = (1/28)*(4 + (54-39*sqrt(2))*(2-2*sqrt(2))^n + (2*(1+sqrt(2)))^n*(54+39*sqrt(2))). - Colin Barker, Nov 13 2017
From Philippe Deléham, Mar 18 2024: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 1.
a(n-1) = Sum_{k = 0..n} A370174(n,k)*3^k. (End)
EXAMPLE
a(0) = [r] = 4, where r = 2+sqrt(8).
a(1) = [4*r] = 19; a(2) = [19*r] = 91.
MATHEMATICA
x = 2 + 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}] (* this sequence *)
t2 = Table[p2[n], {n, 0, z}] (* A057087 *)
t3 = Table[p3[n], {n, 0, z}] (* A086347 *)
t4 = Table[p4[n], {n, 0, z}] (* A218989 *)
PROG
(PARI) Vec((4 - x - 4*x^2) / ((1 - x)*(1 - 4*x - 4*x^2)) + O(x^40)) \\ Colin Barker, Nov 13 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 11 2012
STATUS
approved