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, 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

Cf. A003957, A125578, A125579, A177413.

Sequence in context: A079377 A186379 A186381 * A186153 A215001 A187318

Adjacent sequences: A208669 A208670 A208671 * A208673 A208674 A208675

Keyword

nonn

Author

Ben Branman, Feb 29 2012

Status

approved