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(7), and the limit p1(r) = 3.83798607113023840500712572585708...
See A218987 for the power floor function, p4. For comparison with p1, limit(p4(r)/p1(r) = 4-sqrt(7).
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (5,-1,-3).
FORMULA
a(n) = [x*a(n-1)], where x=2+sqrt(7), a(0) = [x].
a(n) = 5*a(n-1) - a(n-2) - 3*a(n-3).
G.f.: (4 - 2*x - 3*x^2)/(1 - 5*x + x^2 + 3*x^3).
a(n) = (14+(161-61*sqrt(7))*(2-sqrt(7))^n+(2+sqrt(7))^n*(161+61*sqrt(7)))/84. - Colin Barker, Sep 02 2016
EXAMPLE
a(0) = [r] = 4, where r = 2+sqrt(7);
a(1) = [4*r] = 18; a(2) = [18*r] = 83.
MATHEMATICA
x = 2 + Sqrt[7]; 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}] (* A218986 *)
t2 = Table[p2[n], {n, 0, z}] (* A015530 *)
t3 = Table[p3[n], {n, 0, z}] (* A126473 *)
t4 = Table[p4[n], {n, 0, z}] (* A218987 *)
LinearRecurrence[{5, -1, -3}, {4, 18, 83}, 30] (* Harvey P. Dale, Jun 18 2014 *)
PROG
(PARI) a(n) = round((14+(161-61*sqrt(7))*(2-sqrt(7))^n+(2+sqrt(7))^n*(161+61*sqrt(7)))/84) \\ Colin Barker, Sep 02 2016
(PARI) Vec((4-2*x-3*x^2)/((1-x)*(1-4*x-3*x^2)) + O(x^30)) \\ Colin Barker, Sep 02 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 11 2012
STATUS
approved