login
A208672
a(n) = floor[1/(1-f(n))], where f(n) is the least nonnegative number such that f(n)^n = cos(f(n)).
1
1, 3, 5, 7, 9, 10, 12, 14, 15, 17, 19, 20, 22, 23, 25, 27, 28, 30, 32, 33, 35, 37, 38, 40, 41, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 89, 90, 92
OFFSET
0,2
COMMENTS
For n=0, the only possible solution is f(0)=0, which yields a(0)=1.
f(n)->1 as n->infinity.
a(n) ~ -n/log(cos(1))
f(1) = the Dottie number 0.73908513321516 = A003957
f(2) is A125578
f(3) is A125579
a(n) is defined for negative values of n as well.
If we let a(n)=floor[c(n)], c(n)=1/(1-f(n)), then f(n)^n=cos(f(n)) <=> 1-1/c(n) = cos(1-1/c(n))^(1/n) = exp(log(cos(1-1/c(n)))/n) = exp(log(cos(1)+O(1/c(n)^2))/n) = 1+log(cos(1))/n+o(1/n), assuming c(n) ~ c*n, which then yields c = -1/log(cos(1)). - M. F. Hasler, Mar 05 2012
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Dottie Number
Eric Weisstein's World of Mathematics, Cosine
Eric Weisstein's World of Mathematics, Fixed Point
EXAMPLE
For n=4, the only positive solution to x^4=cos(x) is x=0.890553, so a(4)=floor(1/(1-.890553)) = floor(9.13682) = 9, so a(4)=9.
MATHEMATICA
f[n_] := 1/(1 - FindRoot[x^n == Cos[x], {x, 0, 1}, WorkingPrecision -> 1000][[1, 2]]); Table[Floor[f[n]], {n, 0, 100}]
PROG
(PARI) a(n)=1\(1-solve(x=0, 1, x^n-cos(x))) \\ Charles R Greathouse IV, Mar 04, 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Ben Branman, Feb 29 2012
STATUS
approved