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A208672 a(n) = floor[1/(1-f(n))], where f(n) is the least nonnegative number such that f(n)^n = cos(f(n)).

%I A208672 #32 Jun 28 2025 18:58:01
%S A208672 1,3,5,7,9,10,12,14,15,17,19,20,22,23,25,27,28,30,32,33,35,37,38,40,
%T A208672 41,43,45,46,48,50,51,53,54,56,58,59,61,63,64,66,67,69,71,72,74,76,77,
%U A208672 79,80,82,84,85,87,89,90,92
%N A208672 a(n) = floor[1/(1-f(n))], where f(n) is the least nonnegative number such that f(n)^n = cos(f(n)).
%C A208672 For n=0, the only possible solution is f(0)=0, which yields a(0)=1.
%C A208672 f(n)->1 as n->infinity.
%C A208672 a(n) ~ -n/log(cos(1))
%C A208672 f(1) = the Dottie number 0.73908513321516 = A003957
%C A208672 f(2) is A125578
%C A208672 f(3) is A125579
%C A208672 a(n) is defined for negative values of n as well.
%C A208672 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
%H A208672 Charles R Greathouse IV, <a href="/A208672/b208672.txt">Table of n, a(n) for n = 0..10000</a>
%H A208672 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DottieNumber.html">Dottie Number</a>
%H A208672 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Cosine.html">Cosine</a>
%H A208672 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FixedPoint.html">Fixed Point</a>
%e A208672 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.
%t A208672 f[n_] := 1/(1 - FindRoot[x^n == Cos[x], {x, 0, 1}, WorkingPrecision -> 1000][[1,2]]); Table[Floor[f[n]], {n, 0, 100}]
%o A208672 (PARI) a(n)=1\(1-solve(x=0,1,x^n-cos(x))) \\ _Charles R Greathouse IV_, Mar 04 2012
%Y A208672 Cf. A003957, A125578, A125579, A177413.
%K A208672 nonn
%O A208672 0,2
%A A208672 _Ben Branman_, Feb 29 2012