A255676 E.g.f. satisfies: A'(x) = (1 + A(x))*(1 + A(x)^2).

1, 1, 3, 15, 81, 561, 4683, 44415, 479241, 5793921, 77332563, 1130944815, 17984844801, 308888337681, 5698762943643, 112401325405215, 2360158641832761, 52564270139375841, 1237645528139173923, 30717272450961249615, 801500394828539219121

Offset

1,3

Links

Table of n, a(n) for n=1..21.

Formula

E.g.f.: Series_Reversion( Integral 1/((1+x)*(1+x^2)) dx ).

E.g.f.: Series_Reversion( (log((1+x)^2/(1+x^2)) + 2*atan(x))/4 ).

E.g.f. A(x) satisfies: (1 + A(x))^2/(1 + A(x)^2) = exp(4*x) / exp(2*atan(A(x))).

a(n) ~ 2^(2*n+1) * n^n / (exp(n) * Pi^(n+1/2)). - Vaclav Kotesovec, Jul 17 2015

Example

E.g.f.: A(x) = x + x^2/2! + 3*x^3/3! + 15*x^4/4! + 81*x^5/5! + 561*x^6/6! +..
where
(1 + A(x))*(1 + A(x)^2) = 1 + x + 3*x^2/2! + 15*x^3/3! + 81*x^4/4! + 561*x^5/5! +...+ a(n+1)*x^n/n! +...
The series reversion of the e.g.f. equals Integral 1/(1+x+x^2+x^3) dx:
Series_Reversion(A(x)) = x - x^2/2 + x^5/5 - x^6/6 + x^9/9 - x^10/10 + x^13/13 - x^14/16 + x^17/17 - x^18/18 +...
which equals (log((1+x)^2/(1+x^2)) + 2*atan(x))/4.

Mathematica

Rest[CoefficientList[InverseSeries[Series[(Log[(1+x)^2/(1+x^2)] + 2*ArcTan[x])/4, {x, 0, 20}], x], x] * Range[0, 20]!] (* Vaclav Kotesovec, Jul 17 2015 *)

Prog

(PARI) {a(n) = local(A=x); for(i=1, n, A = intformal( (1+A)*(1+A^2) +x*O(x^n))); n!*polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))
(PARI) {a(n) = local(A=x); A = serreverse( intformal( 1/((1+x)*(1+x^2) +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

Crossrefs

Sequence in context: A122868 A264225 A343975 * A015680 A084208 A059271

Adjacent sequences: A255673 A255674 A255675 * A255677 A255678 A255679

Keyword

nonn

Author

Paul D. Hanna, Jul 10 2015

Status

approved