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Define the generalized Pisot 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). This is T(16,32).
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%I #34 Jul 13 2023 09:31:37

%S 16,32,63,124,244,480,944,1856,3649,7174,14104,27728,54512,107168,

%T 210687,414200,814296,1600864,3147216,6187264,12163841,23913482,

%U 47012668,92424472,181701728,357216192,702268543,1380623604,2714234540,5336044608,10490387488

%N Define the generalized Pisot 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). This is T(16,32).

%C Not to be confused with the Pisot T(16,32) sequence, which is essentially A000079. - _R. J. Mathar_, Feb 13 2016

%H Colin Barker, <a href="/A018923/b018923.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.: -(8*x^5+4*x^4+2*x^3+x^2-16) / ((x-1)*(x^5+x^4+x^3+x^2+x-1)). - _Colin Barker_, Dec 21 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 = 2 Last@ s; While[k/s[[i - 1]] >= s[[i - 1]]/s[[i - 2]], k--]; AppendTo[s, k], {i, 3, n}]; s]; T[16, 32, 23] (* or *)

%t a = {16, 32}; Do[AppendTo[a, Ceiling[a[[n - 1]]^2/a[[n - 2]]] - 1], {n, 3, 23}]; a (* _Michael De Vlieger_, Feb 15 2016 *)

%t RecurrenceTable[{a[1] == 16, a[2] == 32, a[n] == Ceiling[a[n-1]^2/a[n-2] - 1]}, a, {n, 40}] (* _Vincenzo Librandi_, Feb 17 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(16, 32, 30) \\ _Colin Barker_, Feb 16 2016

%Y Is this the same sequence as A001949?

%K nonn

%O 0,1

%A _R. K. Guy_