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%I #47 Dec 06 2022 18:39:51
%S 6,36,221,1361,8386,51676,318441,1962321,12092366,74516516,459191461,
%T 2829665281,17437183146,107452764156,662153768081,4080375372641,
%U 25144406003926,154946811396196,954825274381101,5883898457682801
%N Power floor sequence of 3+sqrt(10).
%C See A214992 for a discussion of power floor sequence and the power floor function, p1(x) = lim_{n->oo} a(n,x)/x^n. The present sequence is a(n,r), where r = 3+sqrt(10), and the limit p1(r) = 5.815421188487681054332319082...
%C See A218992 for the power floor function, p4. For comparison with p1, we have lim_{r->oo} p4(r)/p1(r) = (3+sqrt(10))/5 = 1.23245553....
%H Clark Kimberling, <a href="/A218991/b218991.txt">Table of n, a(n) for n = 0..250</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-5,-1).
%F a(n) = floor(r*a(n-1)), where r=3+sqrt(10), a(0) = floor(r).
%F a(n) = 7*a(n-1) - 5*a(n-2) - a(n-3).
%F G.f.: (6 - 6*x - x^2)/(1 - 7*x + 5*x^2 + x^3).
%F a(n) = ((5+sqrt(10))*(3-sqrt(10))^(n+2) + (5-sqrt(10))*(3+sqrt(10))^(n+2)+2)/12. - _Bruno Berselli_, Nov 22 2012
%e a(0) = floor(r) = 6, where r = 3+sqrt(10);
%e a(1) = floor(6*r) = 36;
%e a(2) = floor(36*r) = 221.
%t x = 3 + Sqrt[10]; z = 30; (* z = # terms in sequences *)
%t f[x_] := Floor[x]; c[x_] := Ceiling[x];
%t p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
%t p1[n_] := f[x*p1[n - 1]]
%t p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
%t p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
%t p4[n_] := c[x*p4[n - 1]]
%t t1 = Table[p1[n], {n, 0, z}] (* A218991 *)
%t t2 = Table[p2[n], {n, 0, z}] (* A005668 *)
%t t3 = Table[p3[n], {n, 0, z}] (* A015451 *)
%t t4 = Table[p4[n], {n, 0, z}] (* A218992 *)
%o (Magma) [IsZero(n) select Floor(r) else Floor(r*Self(n)) where r is 3+Sqrt(10): n in [0..20]]; // _Bruno Berselli_, Nov 22 2012
%Y Cf. A214992, A005668, A015451, A218992.
%Y Cf. A176398 (3+sqrt(10)).
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Nov 12 2012
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