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.

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

Offset

0,5

Links

Table of n, a(n) for n=0..24.

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[0] = 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