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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. 2
 1, 1, 1, 0, -3, -12, -27, 0, 441, 3024, 11529, 0, -442827, -4390848, -23444883, 0, 1636819569, 21224560896, 145703137041, 0, -16106380394643, -257991277243392, -2164638920874507, 0, 347592265948756521 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 LINKS M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 Eric W. Weisstein, Jacobi Elliptic Functions FORMULA 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(n, 0)   = n! = A000142(n), F(n, 1/2) = a(n), F(n, 1)   = 2^n*Euler_{n}(1) = A_{n}(-1) = A155585(n). a(2*n) = A159601(n)*(-1)^floor((n-1)/2). a(2*n+1) = A104203(2*n+1). From Peter Bala, Aug 25 2011: (Start) 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 a(2*n) = D^(2*n)[sqrt(1+sin^2(x))](0) a(2*n+1) = D^(2*n)[sqrt(1-x^4)](0). 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 A(u) := E(u,i) = 1+u+u^2/2!-3*u^4/4!-12*u^5/5!-27*u^6/6!+.... 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 (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): 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. End proof. The generating function A(u) satisfies 1/A(u) = A(-u). Compare entries of this sequence with those of A104203, A159600, A193541 and A193544. (End) MAPLE 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: MATHEMATICA a = 1; a[n_] := a[n] =   Sum[Mod[(k+1)^3, 4, -1] Binomial[n-1, k] a[n-k-1] a[k], {k, 0, n-1}]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jun 24 2019 *) CROSSREFS Cf. A060627, A060628, A104203, A159600, A193541, A193544. Sequence in context: A169678 A294366 A110859 * A345960 A271183 A125614 Adjacent sequences:  A190901 A190902 A190903 * A190905 A190906 A190907 KEYWORD sign AUTHOR Peter Luschny, Jul 26 2011 STATUS approved

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Last modified September 26 00:56 EDT 2021. Contains 347664 sequences. (Running on oeis4.)