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%I #39 Jan 12 2020 14:13:51
%S 1,-1,13,-541,9509,-7231801,1695106117,-567547087381,36760132319047,
%T -151856004814953841,113144789723082206461,-103890621918675777804301,
%U 8866964955352146292017421,-8002609021370033485261033939,47038068678960604511245887564401,-421635069078222570953208470234640901
%N Numerators of coefficients of odd powers of x in expansion of f(x) = x cos (x cos (x cos( ... .
%C f(x) satisfies f(x) = x * cos(f(x)), and the coefficients can be determined from this.
%H Andrew Howroyd, <a href="/A309204/b309204.txt">Table of n, a(n) for n = 1..100</a>
%H W. M. Gosper, <a href="/A309204/a309204.pdf">Material from Bill Gosper's Computers & Math talk, M.I.T., 1989</a>, i+38+1 pages, annotated and scanned, included with the author's permission. (There are many blank pages because about half of the original pages were two-sided, half were one-sided.)
%H W. M. Gosper, <a href="/A309204/a309204_2.png">Superposition of plots -4 < z < 4 of z, cos z, z cos z, cos( z cos z), z cos(z cos z), ...</a>, Jul 29 2019
%F From _Christophe Vignat_, Jan 06 2020: (Start)
%F Numerator of (-1)^n/(2^(2*n+1)*(2*n+1)!)*Sum_{k=0..2*n+1} binomial(2*n+1,k)*(2*k-2*n-1)^(2*n).
%F Numerator of 1/(2*n+1)*(coefficient of t^(2*n) in cos(t)^(2*n+1)).
%F Numerator of 1/(2*n+1)*(residue of (cos(t)/t)^(2*n+1) at t=0). (End)
%e f(x) = x - (1/2)*x^3 + (13/24)*x^5 - (541/720)*x^7 + (9509/8064)*x^9 - (7231801/3628800)*x^11 + (1695106117/479001600)*x^13 - (567547087381/87178291200)*x^15 + ...
%e Coefficients are: 1, -1/2, 13/24, -541/720, 9509/8064, -7231801/3628800, 1695106117/479001600, -567547087381/87178291200, 36760132319047/2988969984000, -151856004814953841/ 6402373705728000, 113144789723082206461/2432902008176640000, ...
%p M := 20;
%p f := add(c[i]*z^(2*i+1),i=0..M);
%p f := series(f,z,2*M+3);
%p f2 := series(z*cos(f)-f,z,2*M+3);
%p for i from 0 to M do e[i]:=coeff(f2,z,2*i+1); od:
%p elis:=[seq(e[i],i=0..M)]; clis:=[seq(c[i],i=0..M)];
%p t1 := solve(elis,clis); t2 := op(t1);
%p t3 := subs(t2,clis);
%p map(numer, t3);
%t seq[n_] := Module[{p, k}, p = 1 + O[x]; For[k = 2, k <= n, k++, p = Cos[x*p]]; p] // CoefficientList[#, x^2] & // Numerator;
%t seq[16] (* _Jean-François Alcover_, Sep 07 2019, from PARI *)
%t Table[1/(2*n + 1)* SeriesCoefficient[Cos[t]^(2*n + 1), {t, 0, 2*n}], {n, 0, 15}] // Numerator; (* _Christophe Vignat_, Jan 06 2020 *)
%t Table[1/(2*n + 1)*Residue[(Cos[z]/z)^(2*n + 1), {z, 0}], {n, 0, 15}] // Numerator; (* _Christophe Vignat_, Jan 06 2020 *)
%o (PARI) \\ here F(n) gives n terms of power series.
%o F(n)={my(p=1+O(x));for(k=2, n, p=cos(x*p)); p}
%o seq(n)={my(v=Vec(F(n))); vector(n, k, numerator(v[2*k-1]))} \\ _Andrew Howroyd_, Aug 17 2019
%Y Cf. A309205.
%K sign,frac
%O 1,3
%A _N. J. A. Sloane_, Jul 28 2019
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