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E.g.f. satisfies: A'(x) = A(x)^7 * A(-x) with A(0) = 1.
5

%I #4 Apr 09 2014 21:01:18

%S 1,1,6,76,1296,30976,872976,30638656,1205016576,55768141696,

%T 2815440120576,161768220568576,9971911862317056,684392034689560576,

%U 49826356469468676096,3976369161704254898176,333879082003664326066176,30374928089785982961811456,2889528935298595311805464576

%N E.g.f. satisfies: A'(x) = A(x)^7 * A(-x) with A(0) = 1.

%C Conjectures: (Start)

%C a(n) == 1 (mod 5) for n>=0;

%C a(n) == 0 (mod 17) for n>=16;

%C a(n) == 0 (mod 19) for n>=17;

%C a(n) == 0 (mod 37) for n>=36.

%C (End)

%F E.g.f.: 1/(1 - 5*Series_Reversion( Integral (1 - 25*x^2)^(1/5) dx ))^(1/5).

%e E.g.f.: A(x) = 1 + x + 6*x^2/2! + 76*x^3/3! + 1296*x^4/4! + 30976*x^5/5! +...

%e Related series.

%e A(x)^7 = 1 + 7*x + 84*x^2/2! + 1498*x^3/3! + 34776*x^4/4! + 1006432*x^5/5! +...

%e Note that 1 - 1/A(x)^5 is an odd function:

%e 1 - 1/A(x)^5 = 5*x + 50*x^3/3! + 11000*x^5/5! + 7460000*x^7/7! + 10335200000*x^9/9! +...

%e where Series_Reversion((1 - 1/A(x)^5)/5) = Integral (1-25*x^2)^(1/5) dx.

%o (PARI) {a(n)=local(A=1); for(i=0, n, A=1+intformal(A^7*subst(A, x, -x) +x*O(x^n) )); n!*polcoeff(A, n)}

%o for(n=0, 20, print1(a(n), ", "))

%o (PARI) {a(n)=local(A=1); A=1/(1-5*serreverse(intformal((1-25*x^2 +x*O(x^n))^(1/5))))^(1/5); n!*polcoeff(A, n)}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A236953, A235371, A236955, A236956, A236957, A236958.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Apr 08 2014