A174631 a(n) = Floor[(alpha^n-beta^n)(alpha-beta)], with alpha = (1 + Sqrt(a0))/2; beta = (1 - Sqrt(a0))/2; a0 = real minimal Pisot root of x^3-x-1=0(1.324717957244746).

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 15, 17, 18, 19, 21, 22, 24, 26, 28, 30, 33, 35, 38, 41, 44, 47, 51, 54, 59, 63, 68, 73, 79, 85, 91, 98, 105, 113, 122, 131, 141, 152, 163, 176, 189, 203, 219, 235, 253, 272, 293, 315, 339, 364, 392, 421, 453, 487, 524, 564, 607, 652, 702, 755, 812, 873, 939, 1010, 1086, 1168, 1256

Offset: 0

Author

Roger L. Bagula, Nov 29 2010

Keywords

• nonn

Comments

Limiting ratio is:1.0754819626288792.

The integer 5 in the Fibonacci Binet formula is replaced by the minimal Pisot real root as a beta integer to design a very low ratio sequence.

Crossrefs

Cf. A000931,A174576

Programs

Mathematica

a0 = x /. NSolve[x^3 - x - 1 == 0, x][[3]]
a = (1 + Sqrt[a0])/2; b = (1 - Sqrt[a0])/2;
f[n_] := Floor[FullSimplify[(a^n - b^n)/(a - b)]]
Table[f[n], {n, 0, 100}]

Formula

a0=1.324717957244746;

alpha=1.0754819626288792;

beta=-0.07548196262887907;

a(n)=Floor[(alpha^n-beta^n)/(alpha-beta)]