OFFSET
0,3
COMMENTS
In general, given E(x) = exp( Sum_{n>=1} L(n)*x^n/n ), if
_ [x^n] 1/F(x)^n = L(n) + [x^n] 1/F(x)^(2*n) for n >= 1
then F(x) satisfies
_ F( x/E(x*F(x)) )^2 = F(x) * E(x*F(x)),
_ F( x*E(x*F(x)^2) ) = F(x)^2 / E(x*F(x)^2).
Further, if we define series B(x) and C(x) by
_ [x^n] x/B(x) = (1/n) * [x^n] x/F(x)^n for n >= 1
_ [x^n] x/C(x)^2 = (1/n) * [x^n] x/F(x)^(2*n) for n >= 1
then B(x) and C(x) satisfy
_ C(x)^2 = E(x) * B(x)
_ B(x) = C(x*B(x)) = F( x/B(x) )
_ C(x) = B(x/C(x)) = F( x/C(x)^2 )
_ F(x) = B(x*F(x)) = C(x*F(x)^2)
_ B(x*F(x)^2) = F( x*E(x*F(x)^2) )
_ C(x*F(x)) = F( x/E(x*F(x)) )
_ B(x*F(x)^2) = F(x)^2 / E(x*A(x)^2)
_ C(x*F(x))^2 = F(x) * E(x*A(x))
_ B(x*B(x)) = B(x)^2 / E(x*B(x))
_ C(x/C(x))^2 = C(x) * E(x/C(x)).
The above may be deduced using Lagrange inversion.
Here, L(n) = 1, E(x) = 1/(1-x), and g.f. A(x) = F(x).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..400
FORMULA
G.f. A(x) = Sum_{n>=0} a(n)*x^n along with related series B(x) and C(x) satisfy the following formulas.
(1.a) A( x*(1 - x*A(x)) )^2 = A(x) / (1 - x*A(x)),
(1.b) A( x/(1 - x*A(x)^2) ) = A(x)^2 * (1 - x*A(x)^2).
(2.a) [x^n] 1/A(x)^n = 1 + [x^n] 1/A(x)^(2*n) for n >= 1.
(2.b) [x^n] x/B(x) = (1/n) * [x^n] x/A(x)^n for n >= 1.
(2.c) [x^n] x/C(x)^2 = (1/n) * [x^n] x/A(x)^(2*n) for n >= 1.
(3.a) B(x) = (1-x) * C(x)^2.
(3.b) B(x) = A( x/B(x) ) is the g.f. of A383554.
(3.c) C(x) = A( x/C(x)^2 ) is the g.f. of A383555.
(4.a) A(x) = B(x*A(x)).
(4.b) B(x) = C(x*B(x)).
(4.c) C(x) = B(x/C(x)).
(5.a) A(x) = C(x*A(x)^2).
(5.b) C(x) = A( x*(1-x)/B(x) ).
(6.a) B(x*A(x)^2) = A( x/(1 - x*A(x)^2) ).
(6.b) C(x*A(x)) = A( x*(1 - x*A(x)) ).
(7.a) B(x*A(x)^2) = A(x)^2 * (1 - x*A(x)^2).
(7.b) C(x*A(x))^2 = A(x) / (1 - x*A(x)).
(8.a) B(x*B(x)) = B(x)^2 * (1 - x*B(x)).
(8.b) C(x/C(x))^2 = C(x) / (1 - x/C(x)).
a(n) ~ c * n! * n^(2*log(2)) / log(2)^n, where c = 0.3629445772961071641913898... = 4 * c_A383555. - Vaclav Kotesovec, Jun 09 2025
EXAMPLE
G.f.: A(x) = 1 + x + 4*x^2 + 25*x^3 + 203*x^4 + 1986*x^5 + 22492*x^6 + 287779*x^7 + 4092708*x^8 + 63950627*x^9 + 1088344063*x^10 + ...
where A(x) = (1 - x*A(x)) * A( x*(1 - x*A(x)) )^2,
also, A(x)^2 = A( x/(1 - x*A(x)^2) ) / (1 - x*A(x)^2).
RELATED SERIES.
B(x) = 1 + x + 3*x^2 + 15*x^3 + 106*x^4 + 960*x^5 + 10458*x^6 + 131608*x^7 + 1864069*x^8 + ... + A383554(n)*x^n + ...
where B(x) = A( x/B(x) )
also, B(x)^2 = B(x*B(x)) / (1 - x*B(x)).
C(x) = 1 + x + 2*x^2 + 8*x^3 + 53*x^4 + 474*x^5 + 5160*x^6 + 65044*x^7 + 923050*x^8 + ... + A383555(n)*x^n + ...
where C(x) = A( x/C(x)^2 )
also, C(x) = (1 - x/C(x)) * C(x/C(x))^2.
RELATED TABLE.
The table of coefficients of x^k in 1/A(x)^n begins
n = 1: [1, (-1), -3, -18, -148, -1488, -17367, ...];
n = 2: [1, (-2),(-5), -30, -251, -2572, -30546, ...];
n = 3: [1, -3, -6, (-37),-318, -3333, -40332, ...];
n = 4: [1, -4, (-6), -40,(-357),-3840, -47394, ...];
n = 5: [1, -5, -5, -40, -375,(-4151),-52290, ...];
n = 6: [1, -6, -3, (-38),-378, -4314,(-55481),...];
n = 7: [1, -7, 0, -35, -371, -4368, -57344, ...];
n = 8: [1, -8, 4, -32,(-358),-4344, -58184, ...];
n = 9: [1, -9, 9, -30, -342, -4266, -58245, ...];
n =10: [1, -10, 15, -30, -325,(-4152),-57720, ...];
n =11: [1, -11, 22, -33, -308, -4015, -56760, ...];
n =12: [1, -12, 30, -40, -291, -3864,(-55482),...];
n =13: [1, -13, 39, -52, -273, -3705, -53976, ...];
n =14: [1, -14, 49, -70, -252, -3542, -52311, ...];
...
in which we see [x^n] 1/A(x)^n = 1 + [x^n] 1/A(x)^(2*n) for n >= 1.
From the above table we also obtain the series 1/B(x) and 1/C(x)^2.
1/B(x) = exp((-1)*x + (-5)*x^2/2 + (-37)*x^3/3 + (-357)*x^4/4 + (-4151)*x^5/5 + (-55481)*x^6/6 + ...) = 1 - x - 2*x^2 - 10*x^3 - 75*x^4 - 719*x^5 - 8192*x^6 + ...
1/C(x)^2 = exp((-2)*x + (-6)*x^2/2 + (-38)*x^3/3 + (-358)*x^4/4 + (-4152)*x^5/5 + (-55482)*x^6/6 + ...) = 1 - 2*x - x^2 - 8*x^3 - 65*x^4 - 644*x^5 - 7473*x^6 + ...
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0); m=#A-1;
A[#A] = (1 + polcoef( 1/Ser(A)^(2*m) - 1/Ser(A)^m, m))/m; ); A[n+1]}
for(n=0, 21, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 30 2025
STATUS
approved