%I #37 Jul 13 2023 09:29:00

%S 3,5,8,12,17,24,33,45,61,82,110,147,196,261,347,461,612,812,1077,1428,

%T 1893,2509,3325,4406,5838,7735,10248,13577,17987,23829,31568,41820,

%U 55401,73392,97225,128797,170621,226026,299422,396651,525452,696077,922107

%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(3,5).

%C Not to be confused with the Pisot T(3,5) sequence, which is A020745. - _R. J. Mathar_, Feb 13 2016

%C Is 1 followed by this sequence equal to A167385? - _Bruno Berselli_, Feb 17 2016

%H Reinhard Zumkeller, <a href="/A018917/b018917.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 Conjecture: a(n)=a(n-1)+a(n-2)-a(n-4). G.f.: (3+2*x-x^3)/(1-x)/(1-x^2-x^3). [_Colin Barker_, Feb 16 2012]

%F Conjecture: a(n) = a(n-1) + A000931(n+8). - _Reinhard Zumkeller_, Dec 30 2012

%t RecurrenceTable[{a[1] == 3, a[2] == 5, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 50}] (* _Bruno Berselli_, 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(3, 5, 60) \\ _Colin Barker_, Feb 14 2016

%o (Magma) Tiv:=[3,5]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..50]]; // _Bruno Berselli_, Feb 17 2016

%K nonn

%O 0,1

%A _R. K. Guy_