cp's OEIS Frontend

A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.

%I A263646 #45 Mar 08 2024 01:33:03
%S A263646 1,1,2,1,3,1,1,4,1,1,5,1,1,1,6,1,1,1,7,1,1,1,1,8,1,1,1,1,9,1,1,1,1,1,
%T A263646 10,1,1,1,1,1,11,1,1,1,1,1,1,12,1,1,1,1,1,1,13,1,1,1,1,1,1,1,14,1,1,1,
%U A263646 1,1,1,1,15,1,1,1,1,1,1,1,1,16,1,1,1,1,1,1,1,1,17,1,1,1,1,1,1,1,1,1
%N A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.
%C A263646 Coefficients for an expansion of the Schwarzian derivative of a power series f(x), with f'(0) = 1, expressed in terms of an expansion of the natural logarithm of the derivative of the function G(x) = log(D f(x)) = Sum_{n >= 1} -F(n) x^n/n.
%F A263646 Schwarzian{f(x)} = S{f(x)} = (D^3 f(x)) / (D f(x)) - (3/2) [(D^2 f(x)) / D f(x)]^2 = D [(D^2 f(x)) / D f(x)] - (1/2) [(D^2 f(x)) / D f(x)]^2 = D^2 log[D f(x)] - (1/2) [D log[D f(x)]]^2.
%F A263646 Then, with G(x) = log[D f(x)], S{f(x)} = D^2 G(x) - (1/2) [D G(x)]^2.
%F A263646 With f'(0) = 1, G(x) = log[D f(x)] = sum[n >= 1, -F(n) * x^n/n], and F(n) as Fn,
%F A263646 S{f(x)} = -[(F2 + F1^2/2) + (2 F3 + F1 F2) x + (3 F4 + F1 F3 + F2^2/2) x^2 + (4 F5 + F1 F4 + F2 F3) x^3 + (5 F6 + F1 F5 + F2 F4 + F3^2/2) x^4 + (6 F7 + F1 F6 + F2 F5 + F3 F4) x^5 + (7 F8 + F1 F7 + F2 F6 + F3 F5 + F4^2/2) x^6 + ...] .
%F A263646 This entry's a(m) are the numerators of the coefficients of the binary partitions in the brackets. For the singular partition of the integer n, the coefficient is (n-1); for the symmetric partition, 1/2; and for the rest, 1.
%F A263646 More symmetrically, x^2 S{f(x)}= - sum{n>=2, x^n [(n-1)F(n) + (1/2) sum(k=1 to n-1, F(n-k) F(k))]}.
%F A263646 With f(x)= c(0) + x + c(2) x^2 + ... , F(n) are given by the Faber polynomials of A263916: F(n) = Faber(n,2c(2),3c(3),..,(n+1)c(n+1)).
%e A263646 Partitions by powers of x^n:
%e A263646 n=0: -(F2 + F1^2/2)
%e A263646 n=1: -(2 F3 + F1 F2)
%e A263646 n=2: -(3 F4 + F1 F3 + F2^2/2) =  -[3 F4 + (F1 F3 + F2 F2 + F3 F1) / 2]
%e A263646 n=3: -(4 F5 + F1 F4 + F2 F3)  =  -[4 F5 + (F1 F4 + F2 F3 + F3 F2 + F4 F1) / 2]
%e A263646 n=4: -(5 F6 + F1 F5 + F2 F4 + F3^2/2)
%e A263646 --------------------
%e A263646 Example series:
%e A263646 f(x)= (1/2) / (1-x)^2 = 1/2 + x + (3/2) x^2 + 2x^3 + (5/2)x^4 + ... .
%e A263646 log(f'(x)) = log(1 + 3x + 6x^2 + 10x^3 + ...) = 3x + 3 x^2/2 + 3 x^3/3  + ... .
%e A263646 Then F(n) = -3 for n>=1, and the Schwarzian derivative series is
%e A263646 S{f(x)} = - [(-3 + 3^2/2) + (-2*3 + 3^2) x + (-3*3 + 3^2 + 3^2/2) x^2 + ...] = -3/2 - 3x - (9/2)x^2 - 6x^3 - (15/2)x^4 - ... .
%e A263646 --------------------
%e A263646 The Schwarzian vanishes if and only if acting on a Moebius, or linear fractional, transformation. This corresponds to F(n) = (-1)^(n+1) 2 * d^n, where d is an arbitrary constant.
%e A263646 --------------------
%e A263646 Example polynomial:
%e A263646 f(x) = (x-x1)(x-x2)/-(x1+x2)
%e A263646 log(f'(x)) = log[1 - 2x/(x1+x2)] = Sum_{n>= 1} -(2/(x1+x2))^n x^n/n.
%e A263646 Then F(n) = (2/(x1+x2))^n, and the Schwarzian derivative series is
%e A263646 S{f(x)} = (Sum_{n>= 0} -6(n+1) 2^n (x/(x1+x2))^n) / (x1+x2)^2 = -6 / (x1+x2-2*x)^2 (cf. A001787 and A085750).
%o A263646 (Python) print(sum(([n]+[1]*((n+1)//2) for n in range(1, 18)), [])) # _Andrey Zabolotskiy_, Mar 07 2024
%Y A263646 Cf. A263916, A001787, A085750, A051340.
%K A263646 nonn,tabf,easy
%O A263646 1,3
%A A263646 _Tom Copeland_, Oct 31 2015
%E A263646 More terms from _Tom Copeland_, Oct 01 2016