

A097600


A Binet like formula using the AkiyamaThurston tile roots for a Minimal Pisot theta0 sequence.


0



1, 0, 1, 2, 2, 3, 4, 5, 7, 10, 13, 18, 23, 31, 41, 55, 73, 97, 129, 170, 226, 299, 397, 526, 696, 923, 1223, 1620, 2146, 2843, 3766, 4989, 6610, 8756, 11599, 15366, 20356, 26966, 35723, 47323, 62689, 83046, 110013, 145736, 193059, 255749, 338796
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OFFSET

1,4


COMMENTS

Let r1 = 0.662358978622373051..0.562279512062301289..*i, r2 = complexconjugate(r1), and r3 = 1.3247179572.. = A060006 be the three roots of the polynomial x^3x1. i is the imaginary unit. Then f(n) = (r3^nr2^nr2^(5*n))/(r3r2r2^5) is a sequence of numbers, approximately f(1) = 1, f(2) = 0.756+0.786*i, f(3) = 1.263+0.017*i, f(4) = 2.1929+0.704*i, f(5) = 2.205+0.6866*i etc. a(n) is floor(Re(f(n)).


LINKS

Table of n, a(n) for n=1..47.
TianXiao He, Peter J.S. Shiue, Zihan Nie, Minghao Chen, Recursive sequences and GirardWaring identities with applications in sequence transformation, Electronic Research Archive (2020) Vol. 28, No. 2, 10491062.


MATHEMATICA

NSolve[x^3x1==0, x] r1=0.662358978622373051`0.562279512062301289` I r2=0.662358978622373051`+0.562279512062301289` I r3=1.32471795724474605` (* Binet like formula for the Minimal Pisot*) f[n_]=(r3^n((r2^n)+(r2^(5*n))))/(r3r2r2^5) a=Table[Floor[Re[f[n]]], {n, 1, 50}]


CROSSREFS

Cf. A001644.
Sequence in context: A282949 A292200 A249576 * A136422 A173674 A018128
Adjacent sequences: A097597 A097598 A097599 * A097601 A097602 A097603


KEYWORD

nonn,uned,less


AUTHOR

Roger L. Bagula, Sep 20 2004


STATUS

approved



