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%I #26 Jan 03 2017 00:20:29
%S 1,0,0,0,0,0,0,1,-1,0,0,0,2,-4,2,0,0,0,0,-7,15,-8,2,-3,1,0,0,0,-20,50,
%T -20,-10,-30,40,-10,0,0,0,0,61,-281,520,-634,594,-304,34,0,11,-1,0,0,
%U 0,182,-504,-378,952,2436,-4956,2296,-168,462,-364,42,0,0,0,0,-547,4915,-16368,29530,-36853,42411,-50908,49144,-27669,5857,668,-138,-43,1,0,0,0,-1640,4750,28120,-84310,-19450,223880,-233510,255040,-430280,369410,-115840,6830,-6110,3280,-170,0,0,0,0,4921,-82461,428240,-1018098,1757028,-4061298,8799414,-11619696,8677464,-4315854,2941956,-2402754,1032888,-121338,-22482,1900,171,-1
%N E.g.f. B = B(x,y) satisfies: A^2 + B^2 + C^2 = 1 + y^2 and A^3 + B^3 + C^3 = 1 + y^3, where functions A = A(x,y) and C = C(x,y) are described by A278885 and A278887, respectively.
%C This triangle is a row reversal of triangle A278887.
%H Paul D. Hanna, <a href="/A278886/b278886.txt">Table of n, a(n) for n = 0..1680 of rows 0..40 of the triangle in flattened form.</a>
%F Given e.g.f. B(x,y) = Sum_{n>=0} x^n/n! * Sum_{k=0..2*n} T(n,k)*y^k, functions A = A(x,y), B = B(x,y), and C = C(x,y) satisfy:
%F (1) A^2 + B^2 + C^2 = 1 + y^2,
%F (2) A^3 + B^3 + C^3 = 1 + y^3,
%F where y is a parameter independent of x.
%F Vector [A,B,C] equals the integration of the cross product specified by:
%F (3) [A,B,C] = [0,1,y] + Integral [A,B,C] X [A^2,B^2,C^2] dx,
%F thus [A',B',C'] is orthogonal to [A,B,C] and [A^2,B^2,C^2].
%F Explicitly,
%F (3.a) A = Integral B*C^2 - B^2*C dx,
%F (3.b) B = 1 + Integral C*A^2 - C^2*A dx,
%F (3.c) C = y + Integral A*B^2 - A^2*B dx.
%F Since [A',B',C'] = [A,B,C] X [A^2,B^2,C^2], then
%F (4) A'^2 + B'^2 + C'^2 = (1+y^2)*(A^4 + B^4 + C^4) - (1+y^3)^2.
%F (5) [A',B',C'] X [A,B,C] = (1+y^2)*[A^2,B^2,C^2] - (1+y^3)*[A,B,C];
%F explicitly,
%F (5.a) B'*C - C'*B = (1+y^2)*A^2 - (1+y^3)*A,
%F (5.b) C'*A - A'*C = (1+y^2)*B^2 - (1+y^3)*B,
%F (5.c) A'*B - B'*A = (1+y^2)*C^2 - (1+y^3)*C.
%F Let D = A^4 + B^4 + C^4, then
%F (6) [A^2,B^2,C^2] X [A',B',C'] = D*[A,B,C] - (1+y^3)*[A^2,B^2,C^2];
%F explicitly,
%F (6.a) B^2*C' - C^2*B' = D*A - (1+y^3)*A^2,
%F (6.b) C^2*A' - A^2*C' = D*B - (1+y^3)*B^2,
%F (6.c) A^2*B' - B^2*A' = D*C - (1+y^3)*C^2.
%F ...
%F G.f. for column 3: x^2*(1 + 2*x + 3*x^2) / ((1 + x^2)*(1 + 9*x^2)).
%e This triangle of coefficients T(n,k) of x^n*y^k/n! in B(x,y), for n>=0, k=0..2*n, begins:
%e 1;
%e 0, 0, 0;
%e 0, 0, 0, 1, -1;
%e 0, 0, 0, 2, -4, 2, 0;
%e 0, 0, 0, -7, 15, -8, 2, -3, 1;
%e 0, 0, 0, -20, 50, -20, -10, -30, 40, -10, 0;
%e 0, 0, 0, 61, -281, 520, -634, 594, -304, 34, 0, 11, -1;
%e 0, 0, 0, 182, -504, -378, 952, 2436, -4956, 2296, -168, 462, -364, 42, 0;
%e 0, 0, 0, -547, 4915, -16368, 29530, -36853, 42411, -50908, 49144, -27669, 5857, 668, -138, -43, 1;
%e 0, 0, 0, -1640, 4750, 28120, -84310, -19450, 223880, -233510, 255040, -430280, 369410, -115840, 6830, -6110, 3280, -170, 0;
%e 0, 0, 0, 4921, -82461, 428240, -1018098, 1757028, -4061298, 8799414, -11619696, 8677464, -4315854, 2941956, -2402754, 1032888, -121338, -22482, 1900, 171, -1; ...
%e where e.g.f. B(x,y) = Sum_{n>=0} x^n/n! * Sum_{k=0..2*n} T(n,k)*y^k.
%e ...
%e E.g.f.: B(x,y) = 1 + (-y^4 + y^3)*x^2/2! + (2*y^5 - 4*y^4 + 2*y^3)*x^3/3! +
%e (y^8 - 3*y^7 + 2*y^6 - 8*y^5 + 15*y^4 - 7*y^3)*x^4/4! +
%e (-10*y^9 + 40*y^8 - 30*y^7 - 10*y^6 - 20*y^5 + 50*y^4 - 20*y^3)*x^5/5! +
%e (-y^12 + 11*y^11 + 34*y^9 - 304*y^8 + 594*y^7 - 634*y^6 + 520*y^5 - 281*y^4 + 61*y^3)*x^6/6! +
%e (42*y^13 - 364*y^12 + 462*y^11 - 168*y^10 + 2296*y^9 - 4956*y^8 + 2436*y^7 + 952*y^6 - 378*y^5 - 504*y^4 + 182*y^3)*x^7/7! +...
%e such that functions A = A(x,y), B = B(x,y), and C = C(x,y) satisfy:
%e (1) A^2 + B^2 + C^2 = 1 + y^2 and
%e (2) A^3 + B^3 + C^3 = 1 + y^3.
%e RELATED SERIES.
%e A(x,y) = (y^2 - y)*x + (-y^6 + 3*y^5 - 2*y^4 + 2*y^3 - 3*y^2 + y)*x^3/3! + (2*y^7 - 8*y^6 + 6*y^5 + 6*y^4 - 8*y^3 + 2*y^2)*x^4/4! + (y^10 - 11*y^9 + 20*y^8 - 44*y^7 + 104*y^6 - 104*y^5 + 44*y^4 - 20*y^3 + 11*y^2 - y)*x^5/5! + (-10*y^11 + 100*y^10 - 150*y^9 - 70*y^8 + 130*y^7 + 130*y^6 - 70*y^5 - 150*y^4 + 100*y^3 - 10*y^2)*x^6/6! + (-y^14 + 43*y^13 - 142*y^12 + 466*y^11 - 2245*y^10 + 5423*y^9 - 7480*y^8 + 7480*y^7 - 5423*y^6 + 2245*y^5 - 466*y^4 + 142*y^3 - 43*y^2 + y)*x^7/7! +...
%e C(x,y) = y + (y^2 - y)*x^2/2! + (-2*y^4 + 4*y^3 - 2*y^2)*x^3/3! + (-7*y^6 + 15*y^5 - 8*y^4 + 2*y^3 - 3*y^2 + y)*x^4/4! + (20*y^8 - 50*y^7 + 20*y^6 + 10*y^5 + 30*y^4 - 40*y^3 + 10*y^2)*x^5/5! + (61*y^10 - 281*y^9 + 520*y^8 - 634*y^7 + 594*y^6 - 304*y^5 + 34*y^4 + 11*y^2 - y)*x^6/6! + (-182*y^12 + 504*y^11 + 378*y^10 - 952*y^9 - 2436*y^8 + 4956*y^7 - 2296*y^6 + 168*y^5 - 462*y^4 + 364*y^3 - 42*y^2)*x^7/7! +...
%e The scalar triple product yields
%e [A',B',C'] * ([A,B,C] X [A^2,B^2,C^2]) = A'^2 + B'^2 + C'^2
%e where
%e A'^2 + B'^2 + C'^2 = (y^4 - 2*y^3 + y^2) + (4*y^7 - 8*y^6 + 8*y^5 - 8*y^4 + 4*y^3)*x^2/2! + (-8*y^9 + 16*y^8 - 8*y^7 + 8*y^5 - 16*y^4 + 8*y^3)*x^3/3! +
%e (-28*y^11 + 124*y^10 - 240*y^9 + 356*y^8 - 424*y^7 + 356*y^6 - 240*y^5 + 124*y^4 - 28*y^3)*x^4/4! + (80*y^13 - 200*y^12 - 120*y^11 + 480*y^10 - 200*y^9 + 200*y^7 - 480*y^6 + 120*y^5 + 200*y^4 - 80*y^3)*x^5/5! + (244*y^15 - 2148*y^14 + 7048*y^13 - 13684*y^12 + 20236*y^11 - 25128*y^10 + 26864*y^9 - 25128*y^8 + 20236*y^7 - 13684*y^6 + 7048*y^5 - 2148*y^4 + 244*y^3)*x^6/6! +...
%e Also, we have the relation
%e A'^2 + B'^2 + C'^2 = (1+y^2)*(A^4 + B^4 + C^4) - (1+y^3)^2
%e where
%e A^4 + B^4 + C^4 = (y^4 + 1) + (4*y^5 - 8*y^4 + 4*y^3)*x^2/2! + (-8*y^7 + 16*y^6 - 16*y^4 + 8*y^3)*x^3/3! + (-28*y^9 + 124*y^8 - 212*y^7 + 232*y^6 - 212*y^5 + 124*y^4 - 28*y^3)*x^4/4! + (80*y^11 - 200*y^10 - 200*y^9 + 680*y^8 - 680*y^6 + 200*y^5 + 200*y^4 - 80*y^3)*x^5/5! + (244*y^13 - 2148*y^12 + 6804*y^11 - 11536*y^10 + 13432*y^9 - 13592*y^8 + 13432*y^7 - 11536*y^6 + 6804*y^5 - 2148*y^4 + 244*y^3)*x^6/6! +...
%o (PARI) {T(n,k) = my(A=x, B=1, C=y); for(i=1, n,
%o A = intformal(B*C^2 - B^2*C +x*O(x^n));
%o B = 1 + intformal(C*A^2 - C^2*A);
%o C = y + intformal(A*B^2 - A^2*B); ); polcoeff( n!*polcoeff(B, n, x), k, y)}
%o for(n=0,10, for(k=0,2*n, print1(T(n,k),", "));print(""))
%Y Cf. A278885 (A(x,y)), A278887 (C(x,y)), A278889 (central terms).
%Y Cf. A278746 (A at y=2), A278747 (B at y=2), A278748 (C at y=2).
%K sign,tabf
%O 0,13
%A _Paul D. Hanna_, Dec 19 2016
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