A019495 Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(4,11).

4, 11, 30, 81, 218, 586, 1575, 4233, 11376, 30572, 82159, 220793, 593356, 1594576, 4285239, 11516085, 30948148, 83169572, 223508615, 600653577, 1614187084, 4337941272, 11657715927, 31328764525, 84192434676, 226257439900, 608040726071, 1634039193249

Offset: 0

Author

R. K. Guy

Keywords

• nonn

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory (Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
• Index entries for Pisot sequences

Crossrefs

See A008776 for definitions of Pisot sequences.

Programs

Magma

Iv:=[4,11]; [n le 2 select Iv[n] else Floor(Self(n-1)^2/Self(n-2)): n in [1..40]]; // Bruno Berselli, Feb 04 2016

Maple

a:= proc(n) option remember;
      `if`(n<2, [4, 11][n+1], ceil(a(n-1)^2/a(n-2))-1)
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 18 2015

Mathematica

a = {4, 11}; Do[AppendTo[a, Floor[a[[n]]^2/a[[n - 1]]]], {n, 2, 27}];
a (* Michael De Vlieger, Sep 18 2015 *)

PARI

T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=floor(a[n-1]^2/a[n-2])); a
T(4, 11, 100) \\ Colin Barker, Sep 18 2015

Python

from itertools import islice
def A019495_gen(): # generator of terms
    a, b = 4, 11
    yield from (a,b)
    while True:
        a, b = b, (b**2-1)//a
        yield b
A019495_list = list(islice(A019495_gen(),30)) # Chai Wah Wu, Dec 06 2023