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Pisot sequence P(5,11), a(0)=5, a(1)=11, a(n+1) is the nearest integer to a(n)^2/a(n-1).
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%I #30 Jul 13 2023 09:42:34

%S 5,11,24,52,113,246,536,1168,2545,5545,12081,26321,57346,124941,

%T 272212,593075,1292147,2815232,6133614,13363453,29115278,63434160,

%U 138205538,301111747,656039443,1429328995,3114113637,6784794668

%N Pisot sequence P(5,11), a(0)=5, a(1)=11, a(n+1) is the nearest integer to a(n)^2/a(n-1).

%H Colin Barker, <a href="/A021008/b021008.txt">Table of n, a(n) for n = 0..1000</a>

%H J.-L. Baril, <a href="https://doi.org/10.37236/665">Classical sequences revisited with permutations avoiding dotted pattern</a>, Electronic Journal of Combinatorics, 18 (2011), #P178.

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

%F Conjecture: G.f.: (2x^3+x^2-4x+5)/(-x^4+2x^2-3x+1). - _Ralf Stephan_, May 12 2004

%F Conjecture: a(0)=5, a(1)=11, a(2)=24, a(3)=52, a(n)=3*a(n-1)-2*a(n-2)+a(n-4). - _Harvey P. Dale_, May 19 2015

%t nxt[{a_,b_}]:={b,Round[b^2/a]}; Transpose[NestList[nxt,{5,11},30]][[1]] (* _Harvey P. Dale_, May 19 2015 *)

%t RecurrenceTable[{a[n] == Ceiling[a[n - 1]^2/a[n - 2] - 1/2], a[0] == 5, a[1] == 11}, a, {n, 0, 27}] (* _Michael De Vlieger_, Aug 08 2016 *)

%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, 5, 11) \\ _Colin Barker_, Aug 08 2016

%K nonn

%O 0,1

%A _R. K. Guy_