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A190904 a(n) = Sum_{k=0..n-1} cos(Pi*k/2)*binomial(n-1,k)*a(n-1-k)*a(k) for n > 0, a(0) = 1.

%I A190904 #22 Apr 01 2025 03:58:31
%S A190904 1,1,1,0,-3,-12,-27,0,441,3024,11529,0,-442827,-4390848,-23444883,0,
%T A190904 1636819569,21224560896,145703137041,0,-16106380394643,
%U A190904 -257991277243392,-2164638920874507,0,347592265948756521
%N A190904 a(n) = Sum_{k=0..n-1} cos(Pi*k/2)*binomial(n-1,k)*a(n-1-k)*a(k) for n > 0, a(0) = 1.
%H A190904 M. Abramowitz and I. A. Stegun, eds., <a href="https://personal.math.ubc.ca/~cbm/aands/toc.htm">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
%H A190904 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/JacobiEllipticFunctions.html">Jacobi Elliptic Functions</a>.
%F A190904 Let F(n,x) = Sum_{k=0..n-1} cos(Pi*k*x)*binomial(n-1,k)*F(n-1-k,x)* F(k,x), then
%F A190904 F(n, 0)   = n! = A000142(n),
%F A190904 F(n, 1/2) = a(n),
%F A190904 F(n, 1)   = 2^n*Euler_{n}(1) = A_{n}(-1) = A155585(n).
%F A190904 a(2*n) = A159601(n)*(-1)^floor((n-1)/2).
%F A190904 a(2*n+1) = A104203(2*n+1).
%F A190904 From _Peter Bala_, Aug 25 2011: (Start)
%F A190904 The sequence entries may be calculated as follows: Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. The coefficients in the expansion of D^n[f](x) in powers of f(x) can be found in A145271. Then we have
%F A190904 a(2*n) = D^(2*n)[sqrt(1+sin^2(x))](0)
%F A190904 a(2*n+1) = D^(2*n)[sqrt(1-x^4)](0).
%F A190904 The generating function involves the Jacobian elliptic functions. Define E(u,k) := cn(i*u,k)-i*sn(i*u,k) = 1+u+u^2/2!+(1+k^2)*u^3/3!+(1+4*k^2)*u^4/4!+..., where cn(u,k) and sn(u,k) are Jacobian elliptic functions of modulus k (see A060627 and A060628). Then the e.g.f. A(u) for this sequence is
%F A190904 A(u) := E(u,i) = 1+u+u^2/2!-3*u^4/4!-12*u^5/5!-27*u^6/6!+....
%F A190904 Proof: Using well-known properties of the Jacobian elliptic functions (see for example Abramowitz and Stegun, Chapter 16) we find the generating function A(u) satisfies the differential equation
%F A190904 (d/du)A(u) = dn(i*u,i)*A(u) = 1/2*(A(i*u)+A(-i*u))*A(u), which leads to a recurrence for the coefficients of A(u):
%F A190904 a(n+1) = sum{k=0..floor(n/2)} (-1)^k*binomial(n,2*k)*a(2*k)*a(n-2*k) with a(0) = 1. This recurrence is equivalent to the defining recurrence for this sequence given above.
%F A190904 End proof.
%F A190904 The generating function A(u) satisfies 1/A(u) = A(-u).
%F A190904 Compare entries of this sequence with those of A104203, A159600, A193541 and A193544.
%F A190904 (End)
%p A190904 A190904 := proc(n) option remember; `if`(n=0,1,add(((1-irem(k,2))*(-1)^ iquo(k,2))*binomial(n-1,k)*A190904(n-1-k)*A190904(k),k=0..n-1)) end:
%t A190904 a[0] = 1;
%t A190904 a[n_] := a[n] =
%t A190904   Sum[Mod[(k+1)^3, 4, -1] Binomial[n-1, k] a[n-k-1] a[k], {k, 0, n-1}];
%t A190904 Table[a[n], {n, 0, 24}] (* _Jean-François Alcover_, Jun 24 2019 *)
%Y A190904 Cf. A060627, A060628, A104203, A159600, A193541, A193544.
%K A190904 sign
%O A190904 0,5
%A A190904 _Peter Luschny_, Jul 26 2011