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A320156
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Decimal expansion of the unique real root of x^3 - 3*x^2 + 8*x - 16 = 0, or equivalently, the unique positive root of x^4*(x + 5) - 4^4 = 0.
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3
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2, 4, 2, 3, 3, 1, 8, 3, 4, 4, 7, 5, 3, 0, 7, 2, 0, 8, 3, 9, 6, 3, 7, 5, 4, 9, 2, 4, 6, 2, 8, 2, 9, 1, 0, 3, 9, 6, 0, 1, 8, 7, 7, 0, 6, 6, 2, 6, 6, 1, 9, 6, 3, 3, 1, 1, 7, 2, 8, 7, 2, 3, 0, 1, 0, 0, 3, 7, 7, 8, 6, 9, 0, 8, 3, 4, 1, 5, 0, 6, 8, 8, 1, 2, 1, 2, 7
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OFFSET
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1,1
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COMMENTS
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Let t_1, t_2, ..., t_n be n real numbers in [-1, 1] such that Sum_{i=1..n} (t_i)^k = 0, then lim sup ((Sum_{i=1..n} (t_i)^m)/n) = (x_0)/k, where k > m > 0 are odd numbers and x_0 is the unique positive root of x^(k-m)*(x + k)^m - m^m*(k - m)^(k-m). x_0 is an algebraic integer of degree k - 2. The upper bound (x_0)/k can be approached by x_1 = x_2 = ... = x_(p^k) = -1, x_(p^k+1) = x_(p^k+2) = ... = x_(p^k+q^k) = p/q, where p/q is a rational arbitrarily close to t_0, t_0 is the unique positive root of (k - m)*x^k + k*x^(k - m) - m = 0. Note that (x_0)/k = ((t_0)^m - (t_0)^k)/((t_0)^k + 1).
Here k = 5, m = 1, so the upper bound is (x_0)/5 = 0.48466366895061441679275..., where x_0 = 2.4233183447530720839637... is the unique positive root to x^4*(x + 5) - 4^4 = 0. The upper bound (x_0)/5 can be approached by t_1 = t_2 = ... = t_(p^5) = -1, t_(p^5+1) = t_(p^5+2) = ... = t_(p^5+q^5) = p/q, where p/q is a rational arbitrarily close to t_0, t_0 = 0.60582958618826802099093... is the unique positive root of 4*x^5 + 5*x^4 - 1 = 0. For example, let p = 60583, q = 100000, t_1 = t_2 = ... = t_(60583^5) = -1, t_(60583^5+1) = t_(60583^5+2) = ... = t_(60583^5+100000^5) = 0.60583, then (Sum_{i=1..60583^5+100000^5} t_i)/(60583^5 + 100000^5) = 0.48466366895009176321695..., very close to (x_0)/5. Note that (x_0)/5 = (t_0 - (t_0)^5)/((t_0)^5 + 1).
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LINKS
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EXAMPLE
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2.4233183447530720839637549246282910396018770662662...
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MAPLE
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evalf(solve(x^3-3*x^2+8*x-16=0, x)[1], 120); # Muniru A Asiru, Oct 07 2018
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MATHEMATICA
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Part[RealDigits[N[Root[x^3 - 3x^2 + 8x - 16 , 1], 100]], 1] (* Stefano Spezia, Oct 07 2018 *)
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PROG
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(PARI) default(realprecision, 100); solve(x=2, 3, x^3 - 3*x^2 + 8*x - 16)
(Sage) (x^3-3*x^2+8*x-16==0).find_root(2, 3, x) # G. C. Greubel, Feb 25 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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