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

%I #33 May 29 2024 12:08:38

%S 1,1,1,2,6,16,39,99,271,764,2157,6128,17658,51534,151635,448962,

%T 1337493,4008040,12072594,36524898,110943633,338218626,1034509917,

%U 3173811240,9763898994,30113782641,93094164244,288415278638,895332445053,2784580242557,8675408291598,27072326322939

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

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

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

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

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

%C (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.

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

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

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

%H Paul D. Hanna, <a href="/A371709/b371709.txt">Table of n, a(n) for n = 1..1000</a>

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

%F (1) A(x)^3 = A( x*A(x)^2*(1 + A(x)) ).

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

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

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

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

%F a(n) ~ c * d^n / n^(3/2), where d = 3.2753449994351908157330968510747739... and c = 0.1559869008021616116037651076359... - _Vaclav Kotesovec_, May 03 2024

%F 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

%e 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 + ...

%e where A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.

%e RELATED SERIES.

%e 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 + ...

%e 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 + ...

%e 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 + ...

%e Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then

%e 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) + ...

%e where A005704 is the number of partitions of 3*n into powers of 3.

%e We can show that g.f. A(x) = A( x*A(x)^2*(1 + A(x)) )^(1/3) satisfies

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

%e by substituting x*A(x)^2*(1 + A(x)) for x in (4) to obtain

%e A(x)^3 = x * A(x)^2*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(3^n))

%e which is equivalent to formula (4).

%e SPECIFIC VALUES.

%e A(3/10) = 0.526165645044542830201162330432965674027415264612114520...

%e A(1/4) = 0.353259384374080248921564026412797625837830114153200664...

%e A(1/5) = 0.255218141344695821239609680309162895225297482063273545...

%e 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...

%e 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...

%e 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...

%o (PARI) /* Using series reversion of x/Product_{n>=0} (1 + x^(3^n)) */

%o {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)}

%o for(n=1,35, print1(a(n),", "))

%o (PARI) /* Using A(x)^3 = A( x*A(x)^2 + x*A(x)^3 ) */

%o {a(n) = my(A=[1],F); for(i=1,n, A = concat(A,0); F = x*Ser(A);

%o A[#A] = polcoeff( subst(F,x, x*F^2 + x*F^3 ) - F^3, #A+2) ); A[n]}

%o for(n=1,35, print1(a(n),", "))

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

%Y Cf. A005704, A038464.

%K nonn

%O 1,4

%A _Paul D. Hanna_, May 02 2024

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Last modified July 19 14:41 EDT 2024. Contains 374410 sequences. (Running on oeis4.)