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%I #25 Feb 23 2015 21:29:03
%S 0,-3,15,-45,0,1260,-11340,56700,0,-3742200,48648600,-340540200,0,
%T 40864824000,-694702008000,6252318072000,0,-1187940433680000,
%U 24946749107280000,-274414240180080000,0,75738330289702080000,-1893458257242552000000,24614957344153176000000,0
%N n-th derivative of arctan(x) at x = 1, n >= 4.
%C d^ny/dx^n = (((-1)^(n-1))*(n-1)!)*sin(n*arctan(1/x)) /(1+x^2)^(n/2) - (proof by recurrence). If n = 1, 2, 3, the values of the derivatives at x=1 are respectively 1/2, -1/2, 1/2.
%C d^ny/dx^n = n!*sum(k=1..n, (binomial(n-1,k-1)*(-1)^(n-k)*x^(n-k)*(1+x^2)^(-n)*(-1)^((k-1)/2)*(1+(-1)^(k-1)))/(2*k)). - _Vladimir Kruchinin_, Apr 22 2011
%H Alois P. Heinz, <a href="/A177146/b177146.txt">Table of n, a(n) for n = 4..400</a>
%F a(n) = (((-1)^(n-1))*(n-1)!)*sin(n*arctan(1))/2^(n/2).
%F a(n) = 2^(-n-1)*n!*sum(k=1..n, (((-1)^(k-1)+1)*(-1)^(n-k+(k-1)/2)*binomial(n-1,k-1))/k). - _Vladimir Kruchinin_, Apr 22 2011
%F abs(a(n)) = abs(integrate(x=0..infty, sin(x)*exp(-x)*x^(n-1))) (see Mathematica code below). - _John M. Campbell_, Jun 21 2011
%F E.g.f.: arctan(x+1). - _Alois P. Heinz_, Feb 14 2015
%e a(5) = -3 because d^5y/dx^5 = 384*x^4/(1 + x^2)^5 - 288*x^2/(1 + x^2)^4 + 24/(1 + x^2)^3, and for x=1 we obtain 384/32 - 288/16 + 24/8 = -3.
%p # First program, with the formula:
%p n0:= 50: T:=array(1..n0+1):for n from 1 to n0 do:T[n]:=(((-1)^(n-1))*(n-1) !)*sin(n*arctan(1)) /(2^(n/2)):od:print(T):
%p # Second program, with the Maple instruction D(f):
%p n0:= 50: T:=array(1..n0+1):f:=x->arctan(x):for n from 1 to n0 do:D(f): T[n]:=(D(f)(1)):f:=D(f):od: print(T):
%p # third Maple program:
%p a:= n-> n!*coeff(series(arctan(x+1), x, n+1), x, n):
%p seq(a(n), n=4..40); # _Alois P. Heinz_, Feb 14 2015
%t Table[Abs[Integrate[Sin[x]*E^(-x)*(x^(n - 1)), {x, 0, Infinity}]], {n, 4, 28}] (* _John M. Campbell_, Jun 21 2011 *)
%o (Maxima)
%o a(n):=2^(-n-1)*n!*sum((((-1)^(k-1)+1)*(-1)^(n-k+(k-1)/2)*binomial(n-1,k-1))/k,k,1,n); /* _Vladimir Kruchinin_, Apr 22 2011 */
%Y Cf. A005359 (n-th derivatives of arctan(x) at x = 0).
%K sign
%O 4,2
%A _Michel Lagneau_, May 03 2010
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