A018919 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(3,9).

3, 9, 26, 75, 216, 622, 1791, 5157, 14849, 42756, 123111, 354484, 1020696, 2938977, 8462447, 24366645, 70160958, 202020427, 581694636, 1674922950, 4822748423, 13886550633, 39984728949, 115131438424, 331507764639, 954538564968

Offset: 0

Author

R. K. Guy

Keywords

• nonn
• easy

Comments

Let M denotes the 4 X 4 matrix = row by row (1,1,1,1)(1,1,1,0)(1,1,0,0)(1,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n))=M^n*A where A is the vector (1,1,1,1) then a(n)=y(n+1). - Benoit Cloitre, Apr 02 2002

Not to be confused with the Pisot T(3,9) sequence, which is A000244. - R. J. Mathar, Feb 13 2016

Links

• Vincenzo Librandi, 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 linear recurrences with constant coefficients, signature (3,0,-1).
• Index entries for Pisot sequences

Crossrefs

Cf. A076264.

Programs

Magma

Tiv:=[3,9]; [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

CoefficientList[Series[- (x^2 - 3)/(x^3 - 3 x + 1), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 16 2013 *)
RecurrenceTable[{a[1] == 3, a[2] == 9, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* Bruno Berselli, Feb 17 2016 *)
LinearRecurrence[{3,0,-1},{3,9,26},30] (* Harvey P. Dale, Feb 06 2019 *)

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, 9, 30) \\ Colin Barker, Feb 14 2016

Formula

For n>1, a(n) = ceiling(a(n-1)^2/a(n-2)) - 1.

For n>2, a(n) = 3*a(n-1) - a(n-3).

G.f.: -(x^2-3) / (x^3-3*x+1). - Colin Barker, Dec 13 2012