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Pisot sequence T(4,9), a(n) = floor(a(n-1)^2/a(n-2)).
2

%I #24 Jul 13 2023 09:38:49

%S 4,9,20,44,96,209,455,990,2154,4686,10194,22176,48241,104942,228287,

%T 496607,1080300,2350043,5112193,11120867,24191904,52626132,114480851,

%U 249037213,541745915,1178493097,2563648273,5576861234,12131688091,26390804748,57409535261

%N Pisot sequence T(4,9), a(n) = floor(a(n-1)^2/a(n-2)).

%C Satisfies the linear recurrence a(n) = 3*a(n-1) - 4*a(n-3) + a(n-6) just for n <= 10 (see A019493).

%H Colin Barker, <a href="/A019492/b019492.txt">Table of n, a(n) for n = 0..1000</a>

%H D. W. Boyd, <a href="https://www.researchgate.net/publication/258834801_Linear_recurrence_relations_for_some_generalized_Pisot_sequences">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>

%t RecurrenceTable[{a[0] == 4, a[1] == 9, a[n] == Floor[a[n - 1]^2/a[n - 2]]}, a, {n, 0, 40}] (* _Bruno Berselli_, Feb 04 2016 *)

%t nxt[{a_,b_}]:={b,Floor[b^2/a]}; NestList[nxt,{4,9},40][[All,1]] (* _Harvey P. Dale_, Aug 22 2017 *)

%o (Magma) Iv:=[4,9]; [n le 2 select Iv[n] else Floor(Self(n-1)^2/Self(n-2)): n in [1..40]]; // _Bruno Berselli_, Feb 04 2016

%o (PARI) pisotT(nmax, a1, a2) = {

%o a=vector(nmax); a[1]=a1; a[2]=a2;

%o for(n=3, nmax, a[n] = floor(a[n-1]^2/a[n-2]));

%o a

%o }

%o pisotT(50, 4, 9) \\ _Colin Barker_, Jul 29 2016

%Y See A008776 for definitions of Pisot sequences.

%Y Cf. A019493.

%K nonn

%O 0,1

%A _R. K. Guy_