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A014001 Pisot sequence E(7,15), a(n)=[ a(n-1)^2/a(n-2)+1/2 ]. 1

%I #22 Jun 28 2023 08:22:47

%S 7,15,32,68,145,309,658,1401,2983,6351,13522,28790,61297,130508,

%T 277866,591608,1259600,2681830,5709918,12157058,25883745,55109407,

%U 117334132,249817577,531889747,1132453154,2411120262,5133546494,10929898447,23270984338,49546545623

%N Pisot sequence E(7,15), a(n)=[ a(n-1)^2/a(n-2)+1/2 ].

%H Colin Barker, <a href="/A014001/b014001.txt">Table of n, a(n) for n = 0..1000</a>

%H D. W. Boyd, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa34/aa3444.pdf">Some integer sequences related to the Pisot sequences</a>, Acta Arithmetica, 34 (1979), 295-305.

%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 Known not to satisfy any linear recurrence.

%F There are linear recurrences which match e.g. the first 21 terms, but after a while they always fail. - _N. J. A. Sloane_, Aug 07 2016

%p PisotE := proc(a0,a1,n)

%p option remember;

%p if n = 0 then

%p a0 ;

%p elif n = 1 then

%p a1;

%p else

%p floor( procname(a0,a1,n-1)^2/procname(a0,a1,n-2)+1/2) ;

%p end if;

%p end proc:

%p A014001 := proc(n)

%p PisotE(7,15,n) ;

%p end proc: # _R. J. Mathar_, Feb 12 2016

%t a[0] = 7; a[1] = 15;

%t a[n_] := a[n] = Floor[a[n-1]^2/a[n-2] + 1/2];

%t a /@ Range[0, 30] (* _Jean-François Alcover_, Apr 03 2020 *)

%o (PARI) pisotE(nmax, a1, a2) = {

%o a=vector(nmax); a[1]=a1; a[2]=a2;

%o for(n=3, nmax, a[n] = floor(a[n-1]^2/a[n-2]+1/2));

%o a

%o }

%o pisotE(50, 7, 15) \\ _Colin Barker_, Jul 27 2016

%K nonn

%O 0,1

%A _Simon Plouffe_, Dec 11 1996

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