A022034 Define the generalized Pisot 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). This is T(6,31).

6, 31, 160, 825, 4253, 21924, 113017, 582596, 3003248, 15481566, 79806558, 411398091, 2120732851, 10932252540, 56355115894, 290507292601, 1497547928268, 7719771085196, 39794963809107, 205140687086569, 1057488120864155, 5451288779669969, 28101071561042234

Offset

0,1

Comments

This coincides with the Pisot T(6,31) sequence as defined in A008776 at least up to n <=18000. - R. J. Mathar, Feb 13 2016

The recurrence 5*a(n-1)+a(n-2)-a(n-3)-a(n-5) starts to fail at n=22. - R. J. Mathar, Feb 13 2016

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1402

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

Maple

a:= proc(n) option remember;
      `if`(n<2, [6, 31][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

RecurrenceTable[{a[1] == 6, a[2] == 31, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* Bruno Berselli, Feb 17 2016 *)

Prog

(PARI) T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a
T(6, 31, 40) \\ Colin Barker, Feb 14 2016
(Magma) Tiv:=[6, 31]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..30]]; // Bruno Berselli, Feb 17 2016

Crossrefs

Sequence in context: A077352 A038223 A334650 * A277669 A047665 A003128

Adjacent sequences: A022031 A022032 A022033 * A022035 A022036 A022037

Keyword

nonn

Author

R. K. Guy

Extensions

Incorrect conjectures deleted by Alois P. Heinz, Sep 18 2015

Status

approved