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

Author

R. K. Guy

Keywords

• nonn
• easy

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

Programs

Mathematica

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

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)

Extensions

More terms from Johannes W. Meijer, Aug 14 2010