A048625 Pisot sequence P(4,6).

4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961, 8407925, 12322413, 18059374

Offset: 0

Author

David W. Wilson

Keywords

• nonn

Comments

Conjecture: satisfies a linear recurrence having signature (1, 0, 1). - Harvey P. Dale, Jun 05 2021

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for Pisot sequences

Crossrefs

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

Programs

Maple

P := proc(a0,a1,n)
    option remember;
    if n = 0 then
        a0 ;
    elif n = 1 then
        a1;
    else
        ceil( procname(a0,a1,n-1)^2/procname(a0,a1,n-2)-1/2) ;
    end if;
end proc:
A048625 := proc(n)
    P(4,6,n) ;
end proc: # R. J. Mathar, Feb 12 2016

Mathematica

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]];
a[n_] := P[4, 6, n];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Oct 25 2023, after R. J. Mathar *)

PARI

pisotP(nmax, a1, a2) = {
  a=vector(nmax); a[1]=a1; a[2]=a2;
  for(n=3, nmax, a[n] = ceil(a[n-1]^2/a[n-2]-1/2));
  a
}
pisotP(50, 4, 6) \\ Colin Barker, Aug 08 2016

Formula

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

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