OFFSET
1,1
COMMENTS
Because the tribonacci constant t = A058265 > 1, with Beatty sequence At(n) := floor(n*t), n >= 1 (with At(0) = 0) given in A158919, has the companion sequence Bt := floor(n*s), n >= 1, (with Bt(0) = 0), with 1/t + 1/s = 1, and At and Bt are complementary, disjoint sequences for the positive integers. Note that Bt is not A172278. The first entries n = 0..161 coincide. A172278(162) = 354 but At(193) = A158919(193) = 354, hence A172278 is not complementary together with At. In fact, Bt(162) = 355, which is not a member of At.
s-1 = 1/(t-1) equals the real root of 2*x^3 - 2*x - 1. See the formulas below. - Wolfdieter Lang, Sep 15 2022
LINKS
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.
Wikipedia, Beatty Sequence
FORMULA
s = t/(t - 1) with the tribonacci constant t = A058265, the real root of the cubic x^3 - x^2 - x - 1.
s = (1 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3)) / (-2 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3)).
From Wolfdieter Lang, Sep 15 2022: (Start)
s = 1 + ((1 + (1/9)*sqrt(33))/4)^(1/3)+(1/3)*((1 + (1/9)*sqrt(33))/4)^(-1/3).
s = 1 + ((1 + (1/9)*sqrt(33))/4)^(1/3) + ((1 - (1/9)*sqrt(33))/4)^(1/3).
s = 1 + (2/3)*sqrt(3)*cosh((1/3)*arccosh((3/4)*sqrt(3))). (End)
From Dimitri Papadopoulos, Nov 07 2023: (Start)
s = 1 + t^2/(t+1). (End)
EXAMPLE
s = 2.191487883953118747061354268227517293474691021874288091009781338617685...
MAPLE
Digits := 120: a := (1/4 + sqrt(33)/36)^(1/3): 1 + a + 1/(3*a): evalf(%)*10^98: ListTools:-Reverse(convert(floor(%), base, 10)); # Peter Luschny, Sep 15 2022
MATHEMATICA
With[{t=x/.Solve[x^3-x^2-x-1==0, x][[1]]}, RealDigits[t/(t-1), 10, 120][[1]]] (* Harvey P. Dale, Sep 12 2021 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Wolfdieter Lang, Sep 07 2018
STATUS
approved