A018918 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(3,6).

3, 6, 11, 20, 36, 64, 113, 199, 350, 615, 1080, 1896, 3328, 5841, 10251, 17990, 31571, 55404, 97228, 170624, 299425, 525455, 922110, 1618191, 2839728, 4983376, 8745216, 15346785, 26931731, 47261894, 82938843, 145547524, 255418100, 448227520, 786584465

Offset: 0

Author

R. K. Guy

Keywords

• nonn

Comments

Not to be confused with the Pisot T(3,6) sequence as defined in A008776 which is A007283. - R. J. Mathar, Feb 17 2016

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 sequqences

Crossrefs

Seems to be A010901(n) - 1 and (see conjecture) A077855(n+1).

Programs

Magma

Tiv:=[3,6]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2)-1): n in [1..40]]; // Bruno Berselli, Feb 17 2016

Mathematica

RecurrenceTable[{a[1] == 3, a[2] == 6, a[n] == Ceiling[a[n-1]^2/a[n-2] - 1]}, a, {n, 40}] (* Vincenzo Librandi, Feb 17 2016 *)

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(3, 6, 50) \\ Colin Barker, Jul 29 2016

Formula

Conjectures from Colin Barker, Dec 21 2012: (Start)

a(n) = 3*a(n-1)-3*a(n-2)+2*a(n-3)-a(n-4).

G.f.: -(x^3-2*x^2+3*x-3) / ((x-1)*(x^3-x^2+2*x-1)). (End)

a(n) = ceiling( a(n-1)^2/a(n-2)-1 ), by definition. - Bruno Berselli, Feb 16 2016