A322230 E.g.f.: S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1, as a triangle of coefficients read by rows.

1, 1, 2, 1, 28, 16, 1, 270, 1032, 272, 1, 2456, 36096, 52736, 7936, 1, 22138, 1035088, 4766048, 3646208, 353792, 1, 199284, 27426960, 319830400, 704357760, 330545664, 22368256, 1, 1793606, 702812568, 18598875760, 93989648000, 120536980224, 38188155904, 1903757312, 1, 16142512, 17753262208, 1002968825344, 10324483102720, 28745874079744, 24060789342208, 5488365862912, 209865342976, 1, 145282674, 445736371872, 51882638754240, 1013356176688128, 5416305638467584, 9498855414644736, 5590122715250688, 961530104709120, 29088885112832

Offset: 0

Author

Paul D. Hanna, Dec 14 2018

Keywords

• nonn
• tabl

Comments

Equals a row reversal of triangle A325220.

Appears to be a row reversal of EG1 triangle A162005, which has other formulas.

Compare to sn(x,k) = Integral cn(x,k)*dn(x,k) dx, where sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions (see triangle A060628).

Compare also to Michael Pawellek's generalized elliptic functions.

Examples

			E.g.f.: S(x,k) = x + (2*k^2 + 1)*x^3/3! + (16*k^4 + 28*k^2 + 1)*x^5/5! + (272*k^6 + 1032*k^4 + 270*k^2 + 1)*x^7/7! + (7936*k^8 + 52736*k^6 + 36096*k^4 + 2456*k^2 + 1)*x^9/9! + (353792*k^10 + 3646208*k^8 + 4766048*k^6 + 1035088*k^4 + 22138*k^2 + 1)*x^11/11! + (22368256*k^12 + 330545664*k^10 + 704357760*k^8 + 319830400*k^6 + 27426960*k^4 + 199284*k^2 + 1)*x^13/13! + ...
such that C(x,k)^2 - S(x,k)^2 = 1.
This triangle of coefficients T(n,j) of x^(2*n+1)*k^(2*j)/(2*n+1)! in e.g.f. S(x,k) begins:
1;
1, 2;
1, 28, 16;
1, 270, 1032, 272;
1, 2456, 36096, 52736, 7936;
1, 22138, 1035088, 4766048, 3646208, 353792;
1, 199284, 27426960, 319830400, 704357760, 330545664, 22368256;
1, 1793606, 702812568, 18598875760, 93989648000, 120536980224, 38188155904, 1903757312;
1, 16142512, 17753262208, 1002968825344, 10324483102720, 28745874079744, 24060789342208, 5488365862912, 209865342976; ...
RELATED SERIES.
The related series C(x,k), where C(x,k)^2 - S(x,k)^2 = 1, starts
C(x,k) = 1 + x^2/2! + (8*k^2 + 1)*x^4/4! + (136*k^4 + 88*k^2 + 1)*x^6/6! + (3968*k^6 + 6240*k^4 + 816*k^2 + 1)*x^8/8! + (176896*k^8 + 513536*k^6 + 195216*k^4 + 7376*k^2 + 1)*x^10/10! + (11184128*k^10 + 51880064*k^8 + 39572864*k^6 + 5352544*k^4 + 66424*k^2 + 1)*x^12/12! + (951878656*k^12 + 6453433344*k^10 + 8258202240*k^8 + 2458228480*k^6 + 139127640*k^4 + 597864*k^2 + 1)*x^14/14! + ...
The related series D(x,k), where D(x,k)^2 - k^2*S(x,k)^2 = 1, starts
D(x,k) = 1 + k^2*x^2/2! + (5*k^4 + 4*k^2)*x^4/4! + (61*k^6 + 148*k^4 + 16*k^2)*x^6/6! + (1385*k^8 + 6744*k^6 + 2832*k^4 + 64*k^2)*x^8/8! + (50521*k^10 + 410456*k^8 + 383856*k^6 + 47936*k^4 + 256*k^2)*x^10/10! + (2702765*k^12 + 32947964*k^10 + 54480944*k^8 + 17142784*k^6 + 780544*k^4 + 1024*k^2)*x^12/12! + (199360981*k^14 + 3402510924*k^12 + 8760740640*k^10 + 5199585280*k^8 + 686711040*k^6 + 12555264*k^4 + 4096*k^2)*x^14/14! + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..860 terms of this triangle read by rows 0..40.

Crossrefs

Cf. A000182 (diagonal), A162006, A162007, A162008, A162009, A162010.

Cf. A322231 (C), A322232 (D).

Cf. A325220 (row reversal), A162005.

Programs

PARI

N=10;
{S=x;C=1;D=1; for(i=1,2*N, S = intformal(C*D^2 +O(x^(2*N+1))); C = 1 + intformal(S*D^2); D = 1 + k^2*intformal(S*C*D));}
for(n=0,N, for(j=0,n, print1( (2*n+1)!*polcoeff(polcoeff(S,2*n+1,x),2*j,k),", ")) ;print(""))

Formula

E.g.f. S = S(x,k) = Sum_{n>=0} Sum_{j=0..n} T(n,j) * x^(2*n+1) * k^(2*j) / (2*n+1)!, along with related series C = C(x,k) and D = D(x,k), satisfies:

(1a) S = Integral C*D^2 dx.

(1b) C = 1 + Integral S*D^2 dx.

(1c) D = 1 + k^2 * Integral S*C*D dx.

(2a) C^2 - S^2 = 1.

(2b) D^2 - k^2*S^2 = 1.

(3a) C + S = exp( Integral D^2 dx ).

(3b) D + k*S = exp( k * Integral C*D dx ).

(4a) S = sinh( Integral D^2 dx ).

(4b) S = sinh( k * Integral C*D dx ) / k.

(4c) C = cosh( Integral D^2 dx ).

(4d) D = cosh( k * Integral C*D dx ).

(5a) d/dx S = C*D^2.

(5b) d/dx C = S*D^2.

(5c) d/dx D = k^2 * S*C*D.

From Paul D. Hanna, Mar 31 2019, Apr 20 2019 (Start):

Given sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions, with i^2 = -1, k' = sqrt(1-k^2), then

(6a) S = -i * sn( i * Integral D dx, k),

(6b) C = cn( i * Integral D dx, k),

(6c) D = dn( i * Integral D dx, k).

(7a) S = sc( Integral D dx, k') = sn(Integral D dx, k')/cn(Integral D dx, k'),

(7b) C = nc( Integral D dx, k') = 1/cn(Integral D dx, k'),

(7c) D = dc( Integral D dx, k') = dn(Integral D dx, k')/cn(Integral D dx, k'). (End)

Row sums equal (2*n+1)!*(2*n)!/(n!^2*4^n) = A079484(n), the product of two consecutive odd double factorials.

Main diagonal equals A000182, the tangent numbers.