A339513 Define R_{1}(x)=1, R_{n+1}(x)=(R_n(x)*2*x/(1+x^2))'; then a(n)=R_{n}(1).

1, 0, -1, 3, -2, -45, 347, -756, -13031, 184245, -810034, -11404503, 264733177, -1931955480, -21453955777, 796153961091, -8688345850874, -69492467459925, 4300450718587619, -65896562313762012, -307002797419794407, 37668399518087366325

Offset

1,4

Comments

Let (R_n) be the sequence of rational functions satisfying: R_1(x) = 1; R_{n+1}(x) = (R_n(x) * 2*x/(1+x^2))'. By definition, a(n) = R_n(1).

Applying [Dominici, Theorem 4.1] proves that the e.g.f. of this sequence is the series reversion of log(1+x)/2 + x^2/4 + x/2.

Links

Luc Rousseau, Table of n, a(n) for n = 1..300

D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions, arXiv:math/0501052 [math.CA], 2005.

Formula

a(n) = (Sum_{k=0..n-1} (-1)^k*A214406(n-1,k))/2^(n-1).

a(n) = Sum_{P partition of n-1} A145271(P) * Product_{p part of P} A090932(p)*A075553(p+3).

E.g.f.: series reversion of log(1+x)/2 + x^2/4 + x/2.

Example

R_1(x) = 1,
  so a(1) = R_1(1) = 1.
R_2(x) = (R_1(x)*2*x/(1+x^2))' = (1 * 2*x/(1+x^2))' = 2*(1-x^2)/(1+x^2)^2,
  so a(2) = R_2(1) = 0.
R_3(x) = (R_2(x)*2*x/(1+x^2))' = (2*(1-x^2)/(1+x^2)^2 * 2*x/(1+x^2))' = 4*(1-8*x^2+3*x^4)/(1+x^2)^4, so a(3) = R_3(1) = -1.

Prog

(PARI)
list_a(nmax)=my(n, r); n=1; r=1; print1(subst(r, x, 1), ", "); while(n<nmax, n++; r=(r*2*x/(1+x^2))'; print1(subst(r, x, 1), ", "))
list_a(50)
(PARI) my(x='x+O('x^33)); Vec(serlaplace(serreverse(log(1+x)/2 + x^2/4 + x/2))) \\ Joerg Arndt, Dec 22 2020

Crossrefs

Cf. A214406, A145271, A090932, A075553.

Sequence in context: A093398 A002680 A009395 * A226769 A331474 A093892

Adjacent sequences: A339510 A339511 A339512 * A339514 A339515 A339516

Keyword

sign

Author

Luc Rousseau, Dec 07 2020

Status

approved