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%I #20 Jul 13 2023 09:37:01
%S 3,15,76,386,1961,9963,50618,257170,1306579,6638211,33726124,
%T 171349094,870556961,4422955527,22471287314,114167721214,580041026803,
%U 2946958993287,14972332829596,76068500060858,386473956154025,1963521282342195,9975874867682426
%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(3,15) (agrees with A019478 only for n <= 23).
%H Alois P. Heinz, <a href="/A019477/b019477.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>
%p a:= proc(n) option remember;
%p `if`(n<2, [3, 15][n+1], floor(a(n-1)^2/a(n-2))+1)
%p end:
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Sep 18 2015
%t S[a_, b_, n_] := Block[{s = {a, b}, k}, Do[k = Last@ s + 1; While[k/s[[i - 1]] <= s[[i - 1]]/s[[i - 2]], k++]; AppendTo[s, k], {i, 3, n}]; s]; S[3, 15, 10] (* _Michael De Vlieger_, Feb 15 2016 *)
%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(3, 15, 40) \\ _Colin Barker_, Feb 15 2016
%K nonn
%O 0,1
%A _R. K. Guy_
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