A372530 Expansion of g.f. A(x) satisfying A(x)^2 = A( x*A(x)/(1 - A(x)) ).

1, 1, 3, 9, 33, 125, 501, 2065, 8739, 37685, 165107, 732681, 3286679, 14878885, 67889851, 311896993, 1441536321, 6698017445, 31269529601, 146601334841, 689945263873, 3258334336349, 15436401872405, 73341269533009, 349381321611505, 1668434132560765, 7985390073708765

Offset: 1

Author

Paul D. Hanna, May 13 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 + 3*x^3 + 9*x^4 + 33*x^5 + 125*x^6 + 501*x^7 + 2065*x^8 + 8739*x^9 + 37685*x^10 + 165107*x^11 + 732681*x^12 + ...
where A( x*A(x)/(1 - A(x)) ) = A(x)^2.
RELATED SERIES.
Let R(x) be the series reversion of g.f. A(x), R(A(x)) = x, then
R(x) = x * Product_{n>=0} (1 - x^(2^n)) = x - x^2 - x^3 + x^4 - x^5 + x^6 + x^7 - x^8 - x^9 + x^10 + x^11 - x^12 + x^13 - x^14 - x^15 + x^16 + ... + (-1)^A010060(n-1) * x^n + ...
thus,
x = A(x) * (1 - A(x)) * (1 - A(x)^2) * (1 - A(x)^4) * (1 - A(x)^8) * (1 - A(x)^16) * ... * (1 - A(x)^(2^n)) * ...
SPECIFIC VALUES.
A(t) = 1/3 at t = (1/3) * Product_{n>=0} (1 - 1/3^(2^n)) = 0.195062471888103139123433255203480726664398592...
A(t) = 1/4 at t = (1/4) * Product_{n>=0} (1 - 1/4^(2^n)) = 0.175091932719784804433277263483089433821043251...
A(1/6) = 0.2285942310240955503097133963953487564542629539800372181...
A(1/7) = 0.1803372891149269875688065840927292319030238580575714990...
A(1/8) = 0.1506715662175837437127190414569072051853697889895576799...
A(1/6)^2 = A(t) at t = (1/6)*A(1/6)/(1 - A(1/6)) = 0.0493891023845...
A(1/7)^2 = A(t) at t = (1/7)*A(1/7)/(1 - A(1/7)) = 0.0314305744685...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..520

Crossrefs

Cf. A373312, A373313, A372531, A371713, A371709, A010060.

Programs

PARI

{a(n) = my(A=[0,1]); for(i=1,n, A = concat(A,0); F=Ser(A);
A[#A] = polcoeff( subst(F,x, x*F/(1 - F) ) - F^2, #A) ); H=A; A[n+1]}
for(n=1,30, print1(a(n),", "))

Formula

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

(1) A(x)^2 = A( x*A(x)/(1 - A(x)) ).

(2) A(x)^4 = A( x*A(x)^3/((1 - A(x))*(1 - A(x)^2)) ).

(3) A(x)^8 = A( x*A(x)^7/((1 - A(x))*(1 - A(x)^2)*(1 - A(x)^4)) ).

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

(5) A(x) = x / Product_{n>=0} (1 - A(x)^(2^n)).

(6) A(x) = x * Product_{n>=0} (1 + A(x)^(2^n))^(n+1). - Paul D. Hanna, Jun 26 2024

(7) A(x) = Series_Reversion( x * Product_{n>=0} (1 - x^(2^n)) ).

(8) x = Sum_{n>=1} (-1)^A010060(n-1) * A(x)^n, where A010060 is the Thue-Morse sequence.

The radius of convergence r and A(r) satisfy 1 = Sum_{n>=0} 2^n * A(r)^(2^n)/(1 - A(r)^(2^n)) and r = A(r) * Product_{n>=0} (1 - A(r)^(2^n)), where r = 0.19736158352631556925015099049581233030702919287488... and A(r) = 0.37298513723316144189484491702105095014110332846051...

Given r and A(r) above, A(r) also satisfies 1 = Sum_{n>=0} (n+1)*2^n * A(r)^(2^n)/(1 + A(r)^(2^n)). - Paul D. Hanna, Jun 26 2024