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Expansion of e.g.f. A(x) satisfying A(x/A(x)) = exp(x*A(x)^2).
2

%I #13 Jan 04 2024 09:07:41

%S 1,1,7,118,3457,150376,8869249,669261160,62084355505,6878901271024,

%T 890797404903841,132568595259161656,22370325575395442473,

%U 4233795107469842535544,890606081738110684972705,206651730919408572588445216,52550877215770005095599441249,14564273590596678338725804835680

%N Expansion of e.g.f. A(x) satisfying A(x/A(x)) = exp(x*A(x)^2).

%C Conjecture: a(n) == 1 (mod 3) for n >= 0.

%C Conjecture: a(2*n) == 1 (mod 2) for n >= 0.

%C Conjecture: a(2*n+1) == 0 (mod 2) for n >= 1.

%H Paul D. Hanna, <a href="/A368631/b368631.txt">Table of n, a(n) for n = 0..200</a>

%F E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.

%F (1) A(x/A(x)) = exp(x*A(x)^2).

%F (2) A(x) = exp(x*B(x)^3) where B(x) = A(x*B(x)) = (1/x)*Series_Reversion(x/A(x)).

%F (3) A(x/C(x)^3) = exp(x) where C(x) = A(x/C(x)^2) = ( x/Series_Reversion(x*A(x)^2) )^(1/2).

%e E.g.f.: A(x) = 1 + x + 7*x^2/2! + 118*x^3/3! + 3457*x^4/4! + 150376*x^5/5! + 8869249*x^6/6! + 669261160*x^7/7! + 62084355505*x^8/8! + ...

%e where A(x/A(x)) = exp(x*A(x)^2) and

%e exp(x*A(x)^2) = 1 + x + 5*x^2/2! + 61*x^3/3! + 1377*x^4/4! + 49001*x^5/5! + 2476273*x^6/6! + 165555909*x^7/7! + ...

%e Also,

%e A(x) = exp(x*B(x)^3) where B(x) = A(x*B(x)) begins

%e B(x) = 1 + x + 9*x^2/2! + 187*x^3/3! + 6461*x^4/4! + 320721*x^5/5! + 21079255*x^6/6! + 1741882717*x^7/7! + ...

%e B(x)^3 = 1 + 3*x + 33*x^2/2! + 729*x^3/3! + 25653*x^4/4! + 1275483*x^5/5! + 83368251*x^6/6! + ...

%e Further,

%e A(x/C(x)^3) = exp(x) where C(x) = A(x/C(x)^2) begins

%e C(x) = 1 + x + 3*x^2/2! + 34*x^3/3! + 809*x^4/4! + 30336*x^5/5! + 1584517*x^6/6! + 107443540*x^7/7! + ...

%e C(x)^2 = 1 + 2*x + 8*x^2/2! + 86*x^3/3! + 1944*x^4/4! + 70802*x^5/5! + 3628996*x^6/6! + ...

%o (PARI) {a(n) = my(A=1+x); for(i=0,n, A = exp( x*((1/x)*serreverse( x/(A + x*O(x^n)) ))^3 )); n!*polcoeff(A,n)}

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

%Y Cf. A368630, A368632, A144682, A144683, A367385.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 02 2024