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Pisot sequence P(4,6).
3

%I #38 Oct 25 2023 08:26:28

%S 4,6,9,13,19,28,41,60,88,129,189,277,406,595,872,1278,1873,2745,4023,

%T 5896,8641,12664,18560,27201,39865,58425,85626,125491,183916,269542,

%U 395033,578949,848491,1243524,1822473,2670964,3914488,5736961,8407925,12322413,18059374

%N Pisot sequence P(4,6).

%C Conjecture: satisfies a linear recurrence having signature (1, 0, 1). - _Harvey P. Dale_, Jun 05 2021

%H Colin Barker, <a href="/A048625/b048625.txt">Table of n, a(n) for n = 0..1000</a>

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

%F a(n) = a(n-1) + a(n-3) (Checked up to n = 48000).

%F G.f.: (conjecture) (( Q(0)-1)/2 -(x+x^2+x^3+2*x^4+3*x^5))/x^6, where Q(k) = 1 + x^3 + (2*k+3)*x - x*(2*k+1 + x^2)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Oct 05 2013

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

%p option remember;

%p if n = 0 then

%p a0 ;

%p elif n = 1 then

%p a1;

%p else

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

%p end if;

%p end proc:

%p A048625 := proc(n)

%p P(4,6,n) ;

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

%t P[a0_, a1_, n_] := P[a0, a1, n] = Switch[n, 0, a0, 1, a1, _, Ceiling[P[a0, a1, n-1]^2/P[a0, a1, n-2] - 1/2]];

%t a[n_] := P[4, 6, n];

%t Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, Oct 25 2023, after _R. J. Mathar_ *)

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

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

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

%o a

%o }

%o pisotP(50, 4, 6) \\ _Colin Barker_, Aug 08 2016

%Y Subsequence of A000930. See A008776 for definitions of Pisot sequences.

%K nonn

%O 0,1

%A _David W. Wilson_