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%I #30 Jul 13 2023 09:52:05
%S 16,36,80,177,391,863,1904,4200,9264,20433,45067,99399,219232,483532,
%T 1066464,2352161,5187855,11442175,25236512,55660880,122763936,
%U 270764385,597189651,1317143239,2905050864,6407291380,14131726000,31168502865,68744297111
%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(16,36).
%C Not to be confused with the Pisot T(16,32) sequence, which is essentially A000079. - _R. J. Mathar_, Feb 13 2016
%C Apparently a(n) = A019489(n+2) = A077852(n+3) (Barker's recurrence) for n >= 0. - _Georg Fischer_, Mar 23 2019
%H Colin Barker, <a href="/A022040/b022040.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.: (16-12*x+4*x^2-7*x^3)/(1-3*x+2*x^2-x^3+x^4). - _Colin Barker_, Feb 16 2012
%F a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - _Bruno Berselli_, Feb 15 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, 36, 30) \\ _Colin Barker_, Feb 16 2016
%Y Cf. A019489, A077852.
%K nonn
%O 0,1
%A _R. K. Guy_
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