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%I #33 Jul 13 2023 09:49:14
%S 6,31,160,825,4253,21924,113017,582596,3003248,15481566,79806558,
%T 411398091,2120732851,10932252540,56355115894,290507292601,
%U 1497547928268,7719771085196,39794963809107,205140687086569,1057488120864155,5451288779669969,28101071561042234
%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(6,31).
%C This coincides with the Pisot T(6,31) sequence as defined in A008776 at least up to n <=18000. - _R. J. Mathar_, Feb 13 2016
%C The recurrence 5*a(n-1)+a(n-2)-a(n-3)-a(n-5) starts to fail at n=22. - _R. J. Mathar_, Feb 13 2016
%H Alois P. Heinz, <a href="/A022034/b022034.txt">Table of n, a(n) for n = 0..1402</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>
%p a:= proc(n) option remember;
%p `if`(n<2, [6, 31][n+1], ceil(a(n-1)^2/a(n-2))-1)
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Sep 18 2015
%t RecurrenceTable[{a[1] == 6, a[2] == 31, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* _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(6, 31, 40) \\ _Colin Barker_, Feb 14 2016
%o (Magma) Tiv:=[6,31]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..30]]; // _Bruno Berselli_, Feb 17 2016
%K nonn
%O 0,1
%A _R. K. Guy_
%E Incorrect conjectures deleted by _Alois P. Heinz_, Sep 18 2015
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