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%I #33 Jul 13 2023 09:39:24
%S 4,10,24,57,135,319,753,1777,4193,9893,23341,55069,129925,306533,
%T 723205,1706261,4025589,9497589,22407701,52866581,124728341,294272085,
%U 694277333,1638011349,3864566869,9117688405,21511399509,50751932757,119739242325,282501283669
%N Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(4,10).
%C Not to be confused with the Pisot T(4,10) sequence, which is A020748. - _R. J. Mathar_, Feb 13 2016
%H Colin Barker, <a href="/A019494/b019494.txt">Table of n, a(n) for n = 0..1000</a>
%H D. W. Boyd, <a href="http://www.researchgate.net/publication/258834801">Linear recurrence relations for some generalized Pisot sequences</a>, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
%H <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>
%F Empirical G.f.: (4-2*x+2*x^2-3*x^3)/(1-3*x+2*x^2-2*x^3+2*x^4). - _Colin Barker_, Feb 04 2012
%F a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - _Bruno Berselli_, Feb 15 2016
%t T[a_, b_, n_] := Block[{s = {a, b}, k}, Do[k = Ceiling[b Last@ s/a]; While[k/s[[i - 1]] >= s[[i - 1]]/s[[i - 2]], k--]; AppendTo[s, k], {i, 3, n}]; s]; T[4, 10, 20] (* or *)
%t a = {4, 10}; Do[AppendTo[a, Ceiling[a[[n - 1]]^2/a[[n - 2]]] - 1], {n, 3, 27}]; a (* _Michael De Vlieger_, Feb 15 2016 *)
%o (PARI) T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a
%o T(4, 10, 30) \\ _Colin Barker_, Feb 16 2016
%K nonn
%O 0,1
%A _R. K. Guy_
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