login
Pisot sequence E(4,14): a(n) = floor(a(n-1)^2/a(n-2)+1/2) for n>1, a(0)=4, a(1)=14.
4

%I #32 Sep 08 2022 08:44:37

%S 4,14,49,172,604,2121,7448,26154,91841,322504,1132488,3976785,

%T 13964668,49037590,172197809,604680724,2123364868,7456295833,

%U 26183134320,91943310482,322863269121,1133749589840,3981215131600,13980224615841,49092217790004,172389637059934

%N Pisot sequence E(4,14): a(n) = floor(a(n-1)^2/a(n-2)+1/2) for n>1, a(0)=4, a(1)=14.

%D Shalosh B. Ekhad, N. J. A. Sloane and Doron Zeilberger, Automated Proof (or Disproof) of Linear Recurrences Satisfied by Pisot Sequences, Preprint, 2016.

%H Colin Barker, <a href="/A010904/b010904.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 S. B. Ekhad, N. J. A. Sloane, D. Zeilberger, <a href="http://arxiv.org/abs/1609.05570">Automated proofs (or disproofs) of linear recurrences satisfied by Pisot Sequences</a>, arXiv:1609.05570 [math.NT] (2016)

%F Theorem: a(0)=4, a(1)=14, a(2)=49; for n>2, a(n) = 4*a(n-1)-2*a(n-2)+a(n-3). Proved using the PtoRv program of Ekhad-Sloane-Zeilberger. (Conjectured by _Harvey P. Dale_, May 02 2012.) - _N. J. A. Sloane_, Sep 09 2016

%t RecurrenceTable[{a[0] == 4, a[1] == 14, a[n] == Floor[a[n-1]^2/a[n-2] + 1/2]}, a ,{n, 30}] (* _Harvey P. Dale_, May 02 2012 *)

%o (Magma) I:=[4, 14]; [n le 2 select I[n] else Floor(Self(n-1)^2/Self(n-2)+1/2): n in [1..25]]; // _Bruno Berselli_, Sep 03 2013

%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, 4, 14) \\ _Colin Barker_, Jul 27 2016

%K nonn

%O 0,1

%A _Simon Plouffe_