A371716 Expansion of g.f. A(x) satisfies A( x*A(x)^3 + x*A(x)^4 ) = A(x)^4.

1, 1, 1, 1, 2, 7, 22, 57, 131, 298, 738, 2003, 5600, 15380, 41224, 109769, 296010, 813333, 2261818, 6307070, 17560050, 48877852, 136457322, 382803675, 1078562370, 3047295816, 8623046992, 24432992884, 69345396556, 197211214852, 561975160288, 1604186098089, 4585779820379

Offset: 1

Author

Paul D. Hanna, May 03 2024

Keywords

• nonn

Comments

Compare to the following identities of the Catalan function C(x) = x + C(x)^2 (A000108):

(1) C(x)^2 = C( x*C(x)*(1 + C(x)) ),

(2) C(x)^4 = C( x*C(x)^3*(1 + C(x))*(1 + C(x)^2) ),

(3) C(x)^8 = C( x*C(x)^7*(1 + C(x))*(1 + C(x)^2)*(1 + C(x)^4) ),

(4) C(x)^(2^n) = C( x*C(x)^(2^n-1)*Product_{k=0..n-1} (1 + C(x)^(2^k)) ) for n > 0.

Examples

			G.f.: A(x) = x + x^2 + x^3 + x^4 + 2*x^5 + 7*x^6 + 22*x^7 + 57*x^8 + 131*x^9 + 298*x^10 + 738*x^11 + 2003*x^12 + 5600*x^13 + 15380*x^14 + ...
where A( x*A(x)^3*(1 + A(x)) ) = A(x)^4.
RELATED SERIES.
Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then
B(x) = x/((1+x)*(1+x^4)*(1+x^16)*(1+x^64)*(1+x^256)*(1+x^1024)*...) = x - x^2 + x^3 - x^4 + x^9 - x^10 + x^11 - x^12 + x^33 - x^34 + ...
We can show that g.f. A(x) = A( x*A(x)^3*(1 + A(x)) )^(1/4) satisfies
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(4^n))
by substituting x*A(x)^3*(1 + A(x)) for x in (4) to obtain
A(x)^4 = x * A(x)^3*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(4^n))
which is equivalent to formula (4).
SPECIFIC VALUES.
A(1/3) = 0.6209428791888803994421374991623399343094...
A(1/4) = 0.3392462304609640143453810140211726768116...
A(1/5) = 0.2512464727722296135954631316870173555867...
A(t) = 1/2 and A(t*3/16) = 1/16 at t = 0.31372070319804379323613829910755157...
A(t) = 1/3 and A(t*4/81) = 1/81 at t = 0.24695121377537689193140239461709572...
A(t) = 1/4 and A(t*5/256) = 1/256 at t = 0.199221789836883544932674834867379...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..1030

Crossrefs

Cf. A371709, A371713.

Programs

PARI

/* Using series reversion of x/Product_{n>=0} (1 + x^(4^n)) */
{a(n) = my(A); A = serreverse( x/prod(k=0, ceil(log(n)/log(4)), (1 + x^(4^k) +x*O(x^n)) ) ); polcoeff(A, n)}
for(n=1, 35, print1(a(n), ", "))

PARI

/* Using A(x)^4 = A( x*A(x)^3 + x*A(x)^4 ) */
{a(n) = my(A=[1], F); for(i=1, n, A = concat(A, 0); F = x*Ser(A);
A[#A] = polcoeff( subst(F, x, x*F^3 + x*F^4 ) - F^4, #A+3) ); A[n]}
for(n=1, 35, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.

(1) A(x)^4 = A( x*A(x)^3*(1 + A(x)) ).

(2) A(x)^16 = A( x*A(x)^15*(1 + A(x))*(1 + A(x)^4) ).

(3) A(x)^64 = A( x*A(x)^63*(1 + A(x))*(1 + A(x)^4)*(1 + A(x)^16) ).

(4) A(x) = x * Product_{n>=0} (1 + A(x)^(4^n)).

(5) A(x) = Series_Reversion( x / Product_{n>=0} (1 + x^(4^n)) ).

The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 4^n * A(r)^(4^n) / (1 + A(r)^(4^n)) and r = A(r) / Product_{n>=0} (1 + A(r)^(4^n)), where r = 0.33394799468036632700505690802809657984166722... and A(r) = 0.64588119033501052326223671937159514208118071...