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%I #17 Jul 13 2023 09:37:28
%S 4,8,17,37,81,178,392,864,1905,4201,9265,20434,45068,99400,219233,
%T 483533,1066465,2352162,5187856,11442176,25236513,55660881,122763937,
%U 270764386,597189652,1317143240,2905050865,6407291381,14131726001,31168502866,68744297112
%N Define the sequence S(a_0,a_1) by a_{n+2} is the least integer such that a_{n+2}/a_{n+1}>a_{n+1}/a_n for n >= 0. This is S(4,8).
%H Colin Barker, <a href="/A019479/b019479.txt">Table of n, a(n) for n = 0..1000</a>
%H D. W. Boyd, <a href="https://www.researchgate.net/profile/David_Boyd7/publication/262181133_Linear_recurrence_relations_for_some_generalized_Pisot_sequences_-_annotated_with_corrections_and_additions/links/00b7d536d49781037f000000.pdf">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 Apparently satisfies a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3) - a(n-4).
%F Empirical G.f.: (4-4*x+x^2-2*x^3)/(1-3*x+2*x^2-x^3+x^4). - _Colin Barker_, Feb 04 2012
%o (PARI) S(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=a[n-1]^2\a[n-2]+1); a
%o S(4, 8, 40) \\ _Colin Barker_, Feb 16 2016
%K nonn
%O 0,1
%A _R. K. Guy_
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