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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)
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%I #9 Mar 12 2014 16:37:15

%S 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,

%T 10,11,11,12,13,14,15,17,18,19,21,22,24,26,28,30,33,35,38,41,44,47,51,

%U 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

%N 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)

%C Limiting ratio is:1.0754819626288792.

%C 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.

%F a0=1.324717957244746;

%F alpha=1.0754819626288792;

%F beta=-0.07548196262887907;

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

%t a0 = x /. NSolve[x^3 - x - 1 == 0, x][[3]]

%t a = (1 + Sqrt[a0])/2; b = (1 - Sqrt[a0])/2;

%t f[n_] := Floor[FullSimplify[(a^n - b^n)/(a - b)]]

%t Table[f[n], {n, 0, 100}]

%Y Cf. A000931,A174576

%K nonn

%O 0,13

%A _Roger L. Bagula_, Nov 29 2010