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%I #41 Nov 15 2014 10:14:07
%S 1,1,-1,4,-19,160,-1405,17920,-231175,3942400,-67260025,1435033600,
%T -30485899675,780658278400,-19861319603125,593300291584000,
%U -17590824363487375,600419895083008000,-20332162643728560625,780545863607910400000,-29725015420426543091875,1267606482499246489600000
%N E.g.f. A(x) satisfies: (A(x)^3 - 6*x)^2 = (2 - A(x)^2)^3.
%H Vaclav Kotesovec, <a href="/A249785/b249785.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f.: (1 + 2*Series_Reversion(G(x)))^(1/2), where G(x) = ((1+2*x)^(3/2) - (1-2*x)^(3/2))/6 = x + Sum_{n>=1} x^(2*n+1)/(2*n+1)! * Product_{k=0..n-1} (4*k-1)*(4*k+1).
%F E.g.f. A = A(x) satisfies:
%F (1) A(x)^2 + A(-x)^2 = 2.
%F (2) A(x)^3 - A(-x)^3 = 6*x.
%F (3) (A(x) - A(-x))/2 = Series_Reversion(x - 2*x^3/3).
%F (4) x = (A(x)^3 - (2 - A(x)^2)^(3/2))/6.
%F Recurrence: 4*a(n) = 8*(9*n^2 - 45*n + 55)*a(n-2) - (3*n-11)*(135*n^3 - 1449*n^2 + 5037*n - 5555)*a(n-4) + 9*(n-5)^2*(3*n-19)*(3*n-17)*(3*n-11)*(3*n-7)*a(n-6). - _Vaclav Kotesovec_, Nov 15 2014
%F a(n) ~ (-1)^(n+1) * 3^(n-1/2) * n^(n-1) / (2^((n-1)/2) * exp(n)). - _Vaclav Kotesovec_, Nov 15 2014
%e E.g.f.: A(x) = 1 + x - x^2/2! + 4*x^3/3! - 19*x^4/4! + 160*x^5/5! - 1405*x^6/6! + 17920*x^7/7! - 231175*x^8/8! + 3942400*x^9/9! - 67260025*x^10/10! +...
%e where
%e A(x)^2 = 1 + 2*x + 2*x^3/3! + 50*x^5/5! + 4130*x^7/7! + 723170*x^9/9! + 219099650*x^11/11! + 102195343250*x^13/13! + 68022601897250*x^15/15! +...
%e A(x)^3 = 1 + 3*x + 3*x^2/2! + 21*x^4/4! + 975*x^6/6! + 117705*x^8/8! + 27118875*x^10/10! + 10188133725*x^12/12! + 5671841550375*x^14/14! +...
%e Thus the coefficients of even powers of x in A(x)^2 equal zero:
%e [1, 2, 0, 2, 0, 50, 0, 4130, 0, 723170, 0, 219099650, 0, 102195343250, 0, ...],
%e while the coefficients of odd powers of x in A(x)^3 equal zero:
%e [1, 3, 3, 0, 21, 0, 975, 0, 117705, 0, 27118875, 0, 10188133725, 0, ...],
%e after a few initial terms.
%e EXPLICIT FORMULA.
%e Let G(x) = ((1+2*x)^(3/2) - (1-2*x)^(3/2))/6, which begins
%e G(x) = x - 1*x^3/3! - 1*15*x^5/5! - 1*15*63*x^7/7! - 1*15*63*143*x^9/9! - 1*15*63*143*255*x^11/11! +...+ [Product_{k=0..n-1} (16*k^2 - 1)]*x^(2*n+1)/(2*n+1)! +...
%e then (A(x)^2 - 1)/2 = Series_Reversion(G(x)).
%e The coefficients in G(x) form double factorials (A001147) that begin:
%e [1, 0, -1, 0, -15, 0, -945, 0, -135135, 0, -34459425, 0, -13749310575, 0, ...].
%o (PARI) /* Explicit formula: */
%o {a(n)=local(A, X=x+x^2*O(x^n), G=((1+2*X)^(3/2) - (1-2*X)^(3/2))/6);
%o A=(1 + 2*serreverse(G))^(1/2); n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) /* Formula using series expansion: */
%o {a(n)=local(A,G=x + sum(m=1,n\2+1,x^(2*m+1)/(2*m+1)!*prod(k=0,m-1,(4*k-1)*(4*k+1)) +x^2*O(x^n)));
%o A=sqrt(1 + 2*serreverse(G)); n!*polcoeff(A,n)}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) /* Alternating zero coefficients in A(x)^2 and A(x)^3: */
%o {a(n)=local(A=[1,1],E=1,M);for(i=1,n,A=concat(A,0);M=#A;
%o E=sum(m=0,M-1,A[m+1]*x^m/m!)+x*O(x^M);
%o A[M]=if(M%2==0,-(M-1)!*Vec(E^3/3)[M],-(M-1)!*Vec(E^2/2)[M]));A[n+1]}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A249786 (dual), A249787, A249788.
%K sign
%O 0,4
%A _Paul D. Hanna_, Nov 13 2014
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