A371709 Expansion of g.f. A(x) satisfying A( x*A(x)^2 + x*A(x)^3 ) = A(x)^3.

1, 1, 1, 2, 6, 16, 39, 99, 271, 764, 2157, 6128, 17658, 51534, 151635, 448962, 1337493, 4008040, 12072594, 36524898, 110943633, 338218626, 1034509917, 3173811240, 9763898994, 30113782641, 93094164244, 288415278638, 895332445053, 2784580242557, 8675408291598, 27072326322939

Offset: 1

Author

Paul D. Hanna, May 02 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.

a(3^n) == 1 (mod 3) for n >= 0.

a(2*3^n) == 1 (mod 3) for n >= 0.

a(n) == 2 (mod 3) iff n is the sum of 2 distinct powers of 3 (A038464).

Examples

			G.f. A(x) = x + x^2 + x^3 + 2*x^4 + 6*x^5 + 16*x^6 + 39*x^7 + 99*x^8 + 271*x^9 + 764*x^10 + 2157*x^11 + 6128*x^12 + 17658*x^13 + 51534*x^14 + 151635*x^15 + 448962*x^16 + ...
where A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.
RELATED SERIES.
A(x)^2 = x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 17*x^6 + 48*x^7 + 126*x^8 + 332*x^9 + 918*x^10 + 2616*x^11 + 7504*x^12 + ...
A(x)^3 = x^3 + 3*x^4 + 6*x^5 + 13*x^6 + 36*x^7 + 105*x^8 + 292*x^9 + 801*x^10 + 2256*x^11 + 6515*x^12 + 18981*x^13 + ...
A(x)^2 + A(x)^3 = x^2 + 3*x^3 + 6*x^4 + 12*x^5 + 30*x^6 + 84*x^7 + 231*x^8 + 624*x^9 + 1719*x^10 + 4872*x^11 + 14019*x^12 + 40599*x^13 + ...
Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then
B(x) * (1+x)/(1+x^3) = x - 2*x^4 + 3*x^7 - 5*x^10 + 7*x^13 - 9*x^16 + 12*x^19 - 15*x^22 + 18*x^25 - 23*x^28 + ... + (-1)^n*A005704(n)*x^(3*n+1) + ...
where A005704 is the number of partitions of 3*n into powers of 3.
We can show that g.f. A(x) = A( x*A(x)^2*(1 + A(x)) )^(1/3) satisfies
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n))
by substituting x*A(x)^2*(1 + A(x)) for x in (4) to obtain
A(x)^3 = x * A(x)^2*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(3^n))
which is equivalent to formula (4).
SPECIFIC VALUES.
A(3/10) = 0.526165645044542830201162330432965674027415264612114520...
A(1/4) = 0.353259384374080248921564026412797625837830114153200664...
A(1/5) = 0.255218141344695821239609680309162895225297482063273545...
A(t) = 1/2 and A(t*3/8) = 1/8 at t = (1/2)/Product_{n>=0} (1 + 1/2^(3^n)) = 0.295718718466711580562679377308518930409875701753934072...
A(t) = 1/3 and A(t*4/27) = 1/27 at t = (1/3)/Product_{n>=0} (1 + 1/3^(3^n)) = 0.241059181496179959557718992589733756750585121455883861...
A(t) = 1/4 and A(t*5/64) = 1/64 at t = (1/4)/Product_{n>=0} (1 + 1/4^(3^n)) = 0.196922325724019432212969925740117827612003158137366017...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A371716, A370440, A370441, A370446, A264228, A198951.

Cf. A005704, A038464.

Programs

PARI

/* Using series reversion of x/Product_{n>=0} (1 + x^(3^n)) */
{a(n) = my(A); A = serreverse( x/prod(k=0,ceil(log(n)/log(3)), (1 + x^(3^k) +x*O(x^n)) ) ); polcoeff(A,n)}
for(n=1,35, print1(a(n),", "))

PARI

/* Using A(x)^3 = A( x*A(x)^2 + x*A(x)^3 ) */
{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^2 + x*F^3 ) - F^3, #A+2) ); 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)^3 = A( x*A(x)^2*(1 + A(x)) ).

(2) A(x)^9 = A( x*A(x)^8*(1 + A(x))*(1 + A(x)^3) ).

(3) A(x)^27 = A( x*A(x)^26*(1 + A(x))*(1 + A(x)^3)*(1 + A(x)^9) ).

(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n)).

(5) A(x) = Series_Reversion( x / Product_{n>=0} (1 + x^(3^n)) ).

a(n) ~ c * d^n / n^(3/2), where d = 3.2753449994351908157330968510747739... and c = 0.1559869008021616116037651076359... - Vaclav Kotesovec, May 03 2024

The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 3^n * A(r)^(3^n) / (1 + A(r)^(3^n)) and r = A(r) / Product_{n>=0} (1 + A(r)^(3^n)), where r = 0.30531134893345362211... = 1/d (d is given above) and A(r) = 0.600582105427215700175254768411726892599... - Paul D. Hanna, May 03 2024