%I #79 Sep 08 2022 08:44:37
%S 4,15,56,209,780,2911,10864,40545,151316,564719,2107560,7865521,
%T 29354524,109552575,408855776,1525870529,5694626340,21252634831,
%U 79315912984,296011017105,1104728155436,4122901604639,15386878263120,57424611447841,214311567528244
%N Pisot sequence E(4,15): a(n) = floor(a(n-1)^2/a(n-2)+1/2) for n>1, a(0)=4, a(1)=15.
%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="/A010905/b010905.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 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-1).
%F a(n) = 4*a(n-1) - a(n-2) for n>=2. (Proved using the PtoRv program of Ekhad-Sloane-Zeilberger.) - _N. J. A. Sloane_, Sep 09 2016
%F This was conjectured by _Colin Barker_, Apr 16 2012, and implies the G.f.: (4-x)/(1-4*x+x^2) and the formula a(n) = ((1+sqrt(3))^(2*n+4)-(1-sqrt(3))^(2*n+4))/(2^(n+3)*sqrt(3)).
%F Partial sums of A079935. - _Erin Pearse_, Dec 13 2018
%t a[0] = 4; a[1] = 15; a[n_] := a[n] = Floor[a[n - 1]^2/a[n - 2] + 1/2]; Table[a[n], {n, 0, 24}] (* _Michael De Vlieger_, Jul 27 2016 *)
%o (Magma) /* By definition: */ [n le 2 select 11*n-7 else Floor(Self(n-1)^2/Self(n-2)+1/2): n in [1..22]]; // _Bruno Berselli_, Apr 16 2012
%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, 15) \\ _Colin Barker_, Jul 27 2016
%o (Sage)
%o @cached_function
%o def A010905(n):
%o if n==0: return 4
%o elif n==1: return 15
%o else: return 4*A010905(n-1) - A010905(n-2)
%o [A010905(n) for n in range(30)] # _G. C. Greubel_, Dec 13 2018
%Y Cf. A010925, A001353, A079935, A195503.
%K nonn
%O 0,1
%A _Simon Plouffe_
%E Edited by _N. J. A. Sloane_, Jul 26 2016 and Sep 09 2016