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A012494
Expansion of e.g.f. arctan(sin(x)) (odd powers only).
5
1, -3, 45, -1743, 125625, -14554683, 2473184805, -579439207623, 179018972217585, -70518070842040563, 34495620120141463965, -20515677772241956573503, 14578232896601243652363945, -12198268199871431840616166443, 11871344562637111570703016357525
OFFSET
0,2
COMMENTS
arctan(sin(x)) = x - 3*x^3/3! + 45*x^5/5! - 1743*x^7/7! + 125625*x^9/9! + ....
Absolute values are coefficients in expansion of
arctanh(arcsinh(x)) = x + 3*x^3/3! + 45*x^5/5! + 1743*x^7/7! + ....
arccot(sin(x)) = Pi/2 - x + 3*x^3/3! - 45*x^5/5! + 1743*x^7/7! - ....
LINKS
FORMULA
a(n) = n!*sum(k=1..ceiling(n/2), (1+(-1)^(n-2*k+1))*2^(1-2*k)*sum(i=0..(2*k-1)/2, (-1)^((n+1)/2-i)*binomial(2*k-1,i)*(2*i-2*k+1)^n/n!)/(2*k-1)), n>0. Vladimir Kruchinin, Feb 25 2011
G.f.: cos(x) /(1 + sin^2(x)) = 1 - 3*x^2/2! + 45*x^4/4! - ... . - Peter Bala, Feb 06 2017
a(n) ~ (-1)^n * (2*n)! / (log(1+sqrt(2)))^(2*n+1). - Vaclav Kotesovec, Aug 17 2018
MAPLE
a:= n-> (t-> t!*coeff(series(arctan(sin(x)), x, t+1), x, t))(2*n+1):
seq(a(n), n=0..20); # Alois P. Heinz, Aug 16 2018
MATHEMATICA
Drop[ Range[0, 25]! CoefficientList[ Series[ ArcTan[ Sin[x]], {x, 0, 25}], x], {1, 25, 2}] (* Or *)
f[n_] := n!Sum[(1 + (-1)^(n - 2k + 1))2^(1 - 2k)Sum[(-1)^((n + 1)/2 - j)Binomial[2k - 1, j]((2j - 2k + 1)^n/n!)/(2k - 1), {j, 0, (2k - 1)/2}], {k, Ceiling[n/2]}]; Table[ f[n], {n, 1, 25, 2}] (* Robert G. Wilson v *)
PROG
(Maxima) a(n):=n!*sum((1+(-1)^(n-2*k+1))*2^(1-2*k)*sum((-1)^((n+1)/2-i)*binomial(2*k-1, i)*(2*i-2*k+1)^n/n!, i, 0, (2*k-1)/2)/(2*k-1), k, 1, ceiling((n)/2)); /* Vladimir Kruchinin, Feb 25 2011 */
(Maxima) a(n):=sum(sum((2*i-2*k-1)^(2*n+1)*binomial(2*k+1, i)*(-1)^(n-i+1), i, 0, k)/(4^k*(2*k+1)), k, 0, n); /* Vladimir Kruchinin, Feb 04 2012 */
CROSSREFS
Bisection of A003704, A013208.
Cf. other sequences with a g.f. of the form cos(x)/(1 - k*sin^2(x)): A000364 (k=1), A001209 (k=1/2), A000281 (k=2), A156134 (k=3), A002437 (k=4).
Sequence in context: A144950 A144951 A079484 * A012780 A072503 A154242
KEYWORD
sign,easy
AUTHOR
Patrick Demichel (patrick.demichel(AT)hp.com)
STATUS
approved