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Pisot sequence E(4,21), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
1

%I #31 Jun 29 2023 15:24:55

%S 4,21,110,576,3016,15792,82688,432960,2267008,11870208,62153216,

%T 325438464,1704017920,8922353664,46718050304,244618887168,

%U 1280841121792,6706571182080,35116062605312,183870090903552,962756295000064,5041057406386176,26395319258316800

%N Pisot sequence E(4,21), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).

%C For n >= 1, a(n-1) is the number of generalized compositions of n when there are i+3 different types of i, (i=1,2,...). - _Milan Janjic_, Sep 24 2010

%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="/A010908/b010908.txt">Table of n, a(n) for n = 0..1000</a>

%H Daniel Birmajer, Juan B. Gil, Michael D. Weiner, <a href="https://arxiv.org/abs/1707.07798">(an + b)-color compositions</a>, arXiv:1707.07798 [math.CO], 2017.

%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)

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6, -4).

%F Conjectures from _Colin Barker_, Jun 04 2016: (Start)

%F a(n) = (((3-sqrt(5))^n*(-9+4*sqrt(5)) + (3+sqrt(5))^n*(9+4*sqrt(5))))/(2*sqrt(5)).

%F G.f.: (4-3*x) / (1-6*x+4*x^2).

%F (End)

%F Theorem: a(n) = 6*a(n-1) - 4*a(n-2) for n >= 2. (Proved using the PtoRv program of Ekhad-Sloane-Zeilberger.) This implies the above conjectures. - _N. J. A. Sloane_, Sep 09 2016

%t RecurrenceTable[{a[1] == 4, a[2] == 21, a[n] == Floor[a[n-1]^2/a[n-2]+1/2]}, a, {n, 40}] (* _Vincenzo Librandi_, Aug 09 2016 *)

%o (PARI) a=vector(30); a[1]=4; a[2]=21; for(n=3, #a, a[n]=floor(a[n-1]^2/a[n-2]+1/2)); a \\ _Colin Barker_, Jun 04 2016

%K nonn

%O 0,1

%A _Simon Plouffe_