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A249787 E.g.f. A(x) satisfies: (A(x)^5 - 10*x)^2 = (2 - A(x)^2)^5. 6

%I #43 Nov 20 2014 20:04:31

%S 1,1,-1,0,-3,96,-621,0,-17703,1161216,-13961241,0,-1125897003,

%T 111588212736,-1962807307461,0,-313129783869903,41521527606214656,

%U -961674168773599281,0,-255030903352871048403,42355944224989145726976,-1217055203596100484316701,0

%N E.g.f. A(x) satisfies: (A(x)^5 - 10*x)^2 = (2 - A(x)^2)^5.

%H Paul D. Hanna, <a href="/A249787/b249787.txt">Table of n, a(n) for n = 0..300</a>

%F E.g.f.: (1 + 2*Series_Reversion(G(x)))^(1/2), where G(x) = ((1+2*x)^(5/2) - (1-2*x)^(5/2))/10 = x + Sum_{n>=1} x^(2*n+1)/(2*n+1)! * Product_{k=0..n-1} (4*n-1)*(4*n-3).

%F E.g.f.: sqrt(1 - S(x)^2) + S(x) where S(x) = Series_Reversion(x - 4*x^5/5) so that S(x) = Sum_{n>=0} (4/5)^n * (5*n)!/n! * x^(4*n+1)/(4*n+1)!.

%F E.g.f. A(x) satisfies:

%F (1) A(x)^2 + A(-x)^2 = 2.

%F (2) A(x)^5 - A(-x)^5 = 10*x.

%F (3) (A(x) - A(-x))/2 = Series_Reversion(x - 4*x^5/5).

%F (4) B0^2 + B1^2 = 1 where B0 = (A(x) + A(-x))/2 and B1 = (A(x) - A(-x))/2.

%F (5) x = (A(x)^5 - (2 - A(x)^2)^(5/2))/10.

%F a(4*n+3) = 0 for n>=0.

%e E.g.f.: A(x) = 1 + x - x^2/2! - 3*x^4/4! + 96*x^5/5! - 621*x^6/6! - 17703*x^8/8! + 1161216*x^9/9! - 13961241*x^10/10! +...

%e where

%e A(x)^2 = 1 + 2*x - 6*x^3/3! + 162*x^5/5! - 12726*x^7/7! + 1931202*x^9/9! - 489966246*x^11/11! + 184547004642*x^13/13! - 97585822900566*x^15/15! +...

%e A(x)^5 = 1 + 5*x + 15*x^2/2! - 195*x^4/4! + 9315*x^6/6! - 1049895*x^8/8! + 207604215*x^10/10! - 64920527595*x^12/12! + 29066153231115*x^14/14! +...

%e Thus the coefficients of even powers of x in A(x)^2 equal zero:

%e [1, 2, 0, -6, 0, 162, 0, -12726, 0, 1931202, 0, -489966246, 0, ...],

%e while the coefficients of odd powers of x in A(x)^5 equal zero:

%e [1, 5, 15, 0, -195, 0, 9315, 0, -1049895, 0, 207604215, 0, ...],

%e after a few initial terms.

%e EXPLICIT FORMULA.

%e Let G(x) = ((1+2*x)^(5/2) - (1-2*x)^(5/2))/10, which begins

%e G(x) = x + 3*x^3/3! + 3*3*x^5/5! + 3*3*35*x^7/7! + 3*3*35*99*x^9/9! + 3*3*35*99*195*x^11/11! +...+ [Product_{k=0..n-1} (4*k-1)*(4*k-3)] * x^(2*n+1)/(2*n+1)! +...

%e then (A(x)^2 - 1)/2 = Series_Reversion(G(x)).

%e ALTERNATE SERIES REVERSION FORMULA.

%e Let S(x) = Series_Reversion(x - 4*x^5/5):

%e S(x) = x + 96*x^5/5! + 1161216*x^9/9! + 111588212736*x^13/13! + 41521527606214656*x^17/17! + 42355944224989145726976*x^21/21! +...+ (4/5)^n * (5*n)!/n! * x^(4*n+1)/(4*n+1)! +...

%e then e.g.f. A(x) = S(x) + sqrt(1 - S(x)^2), where

%e sqrt(1 - S(x)^2) = 1 - x^2/2! - 3*x^4/4! - 621*x^6/6! - 17703*x^8/8! +...

%o (PARI) /* Explicit formula: */

%o {a(n)=local(A, X=x+x^2*O(x^n), S=serreverse(X - 4*x^5/5));

%o A = S + sqrt(1 - S^2); n!*polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%o (PARI) /* Explicit formula: */

%o {a(n)=local(A, X=x+x^2*O(x^n), G=((1+2*X)^(5/2) - (1-2*X)^(5/2))/10);

%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-3)) +x^2*O(x^n)));

%o A=(1 + 2*serreverse(G))^(1/2); n!*polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%o (PARI) /* Alternating zero coefficients in A(x)^2 and A(x)^5: */

%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^5/5)[M], -(M-1)!*Vec(E^2/2)[M])); A[n+1]}

%o for(n=0, 25, print1(a(n), ", "))

%Y Cf. A249788 (dual), A249785, A249786, A249923, A249922.

%K sign

%O 0,5

%A _Paul D. Hanna_, Nov 14 2014

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