A181613 Triangle T(n,m) of the coefficients JacobiNC(x,y) = sum_{n>0} sum_{m=0..n-1} (-1)^m* T(n,m) *x^(2*n) *y^(2*m)/(2*n)!.

1, 5, 4, 61, 76, 16, 1385, 2424, 1104, 64, 50521, 113672, 79728, 16832, 256, 2702765, 7432604, 7052528, 2586112, 264448, 1024, 199360981, 647923188, 775638816, 408850432, 85975296, 4205568, 4096, 19391512145, 72718170544, 105138354912, 72490884224, 23551644928, 2939602944, 67162112, 16384

Offset: 1

Author

R. J. Mathar, Jan 30 2011

Keywords

• tabl
• nonn

Comments

The column m=0 is apparently A000364.

Examples

			The triangle starts in row n=1 as:
1;
5, 4;
61, 76, 16;
1385, 2424, 1104, 64;
50521, 113672, 79728, 16832, 256;
		

References

• M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions, Dover. Section 16.22.
• H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 374.

Links

• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
• P. Bala, A triangle for calculating A181613
• D. Dominici, Nested derivatives: A simple method for Computing series expansions of inverse functions, arXiv:math/0501052v2 [math.CA].
• NIST Digital Library of Mathematical Functions, NIST Handbook of Mathematical Functions, Chapter 22.
• Eric W. Weisstein, Jacobi Elliptic Functions.

Crossrefs

Cf. A060627, A060628, A181612.

Programs

Maple

A181613 := proc(n,m) JacobiNC(z,k) ; coeftayl(%,z=0,2*n) ; (-1)^m*coeftayl(%,k=0,2*m)*(2*n)! ; end proc:
seq( seq(A181613(n,m),m=0..n-1),n=1..10) ;

Mathematica

nmax = 8; se = Series[JacobiNC[x, y], {x, 0, 2*nmax}]; t[n_, m_] := Coefficient[se, x, 2*n]*(2*n)! // Coefficient[#, y, m]& // Abs; Table[t[n, m], {n, 1, nmax}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Jan 10 2014 *)

Formula

From Peter Bala, Aug 23 2011: (Start)

The Taylor expansion of the Jacobian elliptic function cn(u,k) begins

cn(u,k) = 1-u^2/2!+(1+4*k^2)*u^4/4!-(1+44*k^2+16*k^4)*u^6/6!+... - see A060627.

The Taylor expansion of the reciprocal function 1/cn(u,k) can be obtained directly from this by using Jacobi's imaginary transformation

1/cn(u,k) = cn(i*u,sqrt(1-k^2)) [Abramowitz and Stegun, 16.20] to yield

1/cn(u,k) = 1+u^2/2!+(5-4*k^2)*u^4/4!+(61-76*k^2+16*k^4)*u^6/6!+....

The coefficient polynomials R(2*n,k) of this expansion can be calculated as follows (apply [Dominici, Theorem 4.1]):

Let f(x) = sqrt(k^2-cos^2(x)). 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. Then R(2*n,k) = D^(2*n)[f](0).

See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x).

(End)

G.f. 1/(1 - x/(1 - 2^2*(1 - k^2)*x/(1 - 3^2*x/(1 - 4^2*(1 - k^2)*x/(1 - 5^2*x/(1 - ...)))))) = 1 + x + (5 - 4*k^2)*x^2 + (61 - 76*k^2 + 16*k^4)*x^3 + ... (see Wall, 94.19, p. 374).