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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(6,37).
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%I #24 Jul 13 2023 09:49:47

%S 6,37,228,1404,8645,53230,327753,2018073,12425877,76509828,471093813,

%T 2900665005,17860258910,109970936934,677123832923,4169253239949,

%U 25671334745061,158066058755653,973259831585207,5992650839998179,36898537188819414,227195290202341077

%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(6,37).

%C This coincides with the Pisot T(6,37) sequence as defined in A008776 at least up to n<=16000. - _R. J. Mathar_, Feb 13 2016

%H Colin Barker, <a href="/A022035/b022035.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>

%F Empirical g.f.: -(x^6+x^5+x^4+x^3-x-6) / (x^7+x^6+x^5+x^4-x^2-6*x+1). - _Colin Barker_, Sep 18 2015

%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, 37, 30) \\ _Colin Barker_, Feb 16 2016

%K nonn

%O 0,1

%A _R. K. Guy_