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Power ceiling-floor sequence of 2+sqrt(2).
4

%I #32 Oct 28 2024 02:00:53

%S 4,13,45,153,523,1785,6095,20809,71047,242569,828183,2827593,9654007,

%T 32960841,112535351,384219721,1311808183,4478793289,15291556791,

%U 52208640585,178251448759,608588513865,2077851157943,7094227604041,24221208100279,82696377193033

%N Power ceiling-floor sequence of 2+sqrt(2).

%C 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 = 2+sqrt(2), and the limit p3(r) = 3.8478612632206289...

%C a(n) is the number of words over {0,1,2,3} of length n+1 that avoid 23, 32, and 33. As an example, a(2)=45 corresponds to the 45 such words of length 3; these are all 64 words except for the 19 prohibited cases which are 320, 321, 322, 323, 230, 231, 232, 233, 330, 331, 332, 333, 023, 123, 223, 032, 132, 033, 133. - _Greg Dresden_ and _Mina BH Arsanious_, Aug 09 2023

%H Clark Kimberling, <a href="/A214997/b214997.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 (3,2,-2).

%F a(n) = floor(x*a(n-1)) if n is odd, a(n) = ceiling(x*a(n-1)) if n is even, where x = 2+sqrt(2) and a(0) = ceiling(x).

%F a(n) = 3*a(n-1) + 2*a(n-2) - 2*a(n-3).

%F G.f.: (4 + x - 2*x^2)/(1 - 3*x - 2*x^2 + 2*x^3).

%F a(n) = (1/14)*(2*(-1)^n + (27-19*sqrt(2))*(2-sqrt(2))^n + (2+sqrt(2))^n*(27+19*sqrt(2))). - _Colin Barker_, Nov 13 2017

%e a(0) = ceiling(r) = 4, where r = 2+sqrt(2);

%e a(1) = floor(4*r) = 13; a(2) = ceiling(13*r) = 45.

%t (See A214996.)

%t CoefficientList[Series[(4+x-2*x^2)/(1-3*x-2*x^2+2*x^3), {x,0,50}], x] (* _G. C. Greubel_, Feb 01 2018 *)

%o (PARI) Vec((4 + x - 2*x^2) / ((1 + x)*(1 - 4*x + 2*x^2)) + O(x^40)) \\ _Colin Barker_, Nov 13 2017

%o (Magma) Q:=Rationals(); R<x>:=PowerSeriesRing(Q, 40); Coefficients(R!((4 +x-2*x^2)/(1-3*x-2*x^2+2*x^3))) // _G. C. Greubel_, Feb 01 2018

%Y Cf. A214992, A007052, A214996, A007070.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Nov 10 2012