A018916 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(2,8).

2, 8, 31, 120, 464, 1794, 6936, 26816, 103676, 400832, 1549696, 5991432, 23164064, 89556864, 346244592, 1338650240, 5175487232, 20009459744, 77360538496, 299091179520, 1156345798592, 4470662117376, 17284466110464, 66825172844672

Offset

0,1

Comments

From Johannes W. Meijer, Aug 14 2010: (Start)

The sequence b(n+1)=2*a(n), n>= 0 with b(0)=1, is a berserker sequence, see A180141. For the corner squares 16 A[5] vectors, with decimal values between 19 and 400, lead to the b(n) sequence.

(End)

Not to be confused with the Pisot T(2,8) sequence, which is A004171. - R. J. Mathar, Feb 13 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 linear recurrences with constant coefficients, signature (4,0,-2).

Index entries for Pisot sequences

Formula

From Johannes W. Meijer, Aug 14 2010: (Start)

G.f.: (2-x^2)/(1-4*x+2*x^3).

a(n) = 4*a(n-1)-2*a(n-3) with a(0)=2, a(1)=8 and a(2)=31.

a(n) = (119-24*z1-64*z1^2)*z1^(-n-1)/202+(119-24*z2-64*z2^2)*z2^(-n-1)/202+(119-24*z3-64*z3^2)*z3^(-n-1)/202 with alpha=2*arctan(sqrt(303)/9), p=(sqrt(6)/3)*sin((alpha+Pi)/6), q=sqrt(2)*cos((alpha+Pi)/6), z1:=2*p, z2=(-q-p) and z3=(q-p).

(End)

Mathematica

LinearRecurrence[{4, 0, -2}, {2, 8, 31}, 25] (* Vincenzo Librandi, Feb 15 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(2, 8, 30) \\ Colin Barker, Feb 14 2016

Crossrefs

Sequence in context: A077838 A389157 A216318 * A391222 A281831 A206229

Adjacent sequences: A018913 A018914 A018915 * A018917 A018918 A018919

Keyword

nonn,easy

Author

R. K. Guy

Extensions

More terms from Johannes W. Meijer, Aug 14 2010

Status

approved