A281180 - OEIS

A281180

a(n) = (2*n-1)! * [x^(2*n-1)] S(x), where S(x) = Integral (1 + S(x)^2)^2 dx.

1, 4, 88, 4672, 454144, 70084096, 15728822272, 4836914249728, 1952137912385536, 1000749157519458304, 635146072839001735168, 488855521088102855606272, 448599416591747486039670784, 483861305506660094099058589696, 606050665000453965359938841608192, 872366179652871528356910686198038528, 1430068361869553198039835379199635357696, 2648687881942689612933392158083076801429504, 5503854158077547090902251582359116752300802048

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Conjecture: the sequence {a(n)} reduced modulo a positive integer k is eventually periodic with the period dividing phi(k)^2, where phi(n) = A000010(n). An example is given below. - Peter Bala, Feb 18 2026

LINKS

Seiichi Manyama, Table of n, a(n) for n = 1..217 (terms 1..100 from Paul D. Hanna)

Peter Bala, The method of Graves for computing inverse functions.

Diego Dominici, Nested derivatives: a simple method for computing series expansions of inverse functions, Int. Jour. of Mathematics and Mathematical Sciences, Vol. 2003, Issue 58, pp. 3699-3715. arXiv:math/0501052v2 [math.CA], 2005.

FORMULA

E.g.f. S(x) = Series_Reversion( Integral 1/(1 + x^2)^2 dx ).

E.g.f. S(x) = Series_Reversion( ( x/(1+x^2) + atan(x) )/2 ).

E.g.f. S(x) and related series C(x) (e.g.f. of A281181) satisfy:

(1.a) C(x)^2 - S(x)^2 = 1.

(1.b) C(x)^2 + S(x)^2 = 1 + Integral 4*C(x)^4*S(x) dx.

Integrals.

(2.a) S(x) = Integral C(x)^4 dx.

(2.b) C(x) = 1 + Integral C(x)^3*S(x) dx.

Exponential.

(3.a) C(x) + S(x) = exp( Integral C(x)^3 dx ).

(3.b) C(x) = cosh( Integral C(x)^3 dx ).

(3.c) S(x) = sinh( Integral C(x)^3 dx ).

Derivatives.

(4.a) S'(x) = C(x)^4.

(4.b) C'(x) = C(x)^3*S(x).

(4.c) (C'(x) + S'(x))/(C(x) + S(x)) = C(x)^3.

(4.d) (C(x)^2 + S(x)^2)' = 4*C(x)^4*S(x).

Explicit Solutions.

(5.a) S(x) = Series_Reversion( Integral 1/(1 + x^2)^2 dx ).

(5.b) C(x) = d/dx Series_Reversion( Integral sqrt(1 - x^2) dx ).

(5.c) C(x) + S(x) = exp( Series_Reversion( Integral 1/cosh(x)^3 dx ) ).

(5.d) C(x)^2 = d/dx Series_Reversion( Integral cos(x)^2 dx ).

(5.e) C(x)^3 = d/dx Series_Reversion( Integral 1/cosh(x)^3 dx ).

For n >= 2, define f(n, x) = d/dx( (1 + x^2)^2 * f(n-1, x) ) with f(1, x) = 1. Then a(n) = f(2*n-1, 0) (apply Dominici, Theorem 4.1, to S(x) = Series_Reversion( Integral 1/(1 + x^2)^2 dx ). - Peter Bala, Dec 22 2025

From Peter Bala, Jan 26 2026: (Start)

S(x) = tan( Series_Reversion(x/2 + sin(2*x)/4) ).

For n >= 2, define g(n, x) = d/dx( 1/sqrt(1 - x^2) * g(n-1, x) ) with g(1, x) = 1/(1 - x^2)^(3/2). Then a(n) = g(2*n-1, 0).

For n >= 1, define h(n, x) = sec(x)^2 * d/dx( h(n-1, x) ) with h(0, x) = tan(x). Then a(n) = h(2*n-1, 0). (End)

EXAMPLE

E.g.f.: S(x) = x + 4*x^3/3! + 88*x^5/5! + 4672*x^7/7! + 454144*x^9/9! + 70084096*x^11/11! + 15728822272*x^13/13! + 4836914249728*x^15/15! + 1952137912385536*x^17/17! + 1000749157519458304*x^19/19! + 635146072839001735168*x^21/21! +...

such that

(1) C(x)^2 - S(x)^2 = 1, and

(2) S'(x) = C(x)^4,

where C(x) begins:

C(x) = 1 + x^2/2! + 13*x^4/4! + 493*x^6/6! + 37369*x^8/8! + 4732249*x^10/10! + 901188997*x^12/12! + 240798388357*x^14/14! + 85948640603761*x^16/16! + 39504564917358001*x^18/18! + 22726779729476308093*x^20/20! +...+ A281181(n)*x^(2*n)/(2*n)! +...

RELATED SERIES.

As power series with reduced fractional coefficients, S(x) and C(x) begin:

S(x) = x + 2/3*x^3 + 11/15*x^5 + 292/315*x^7 + 3548/2835*x^9 + 273766/155925*x^11 + 15360178/6081075*x^13 + 214706776/58046625*x^15 +...

C(x) = 1 + 1/2*x^2 + 13/24*x^4 + 493/720*x^6 + 37369/40320*x^8 + 4732249/3628800*x^10 + 901188997/479001600*x^12 + 240798388357/87178291200*x^14 +...

The series reversion of the e.g.f. begins:

Series_Reversion(S(x)) = x - 2/3*x^3 + 3/5*x^5 - 4/7*x^7 + 5/9*x^9 - 6/11*x^11 + 7/13*x^13 - 8/15*x^15 +...

which equals ( x/(1+x^2) + atan(x) )/2.

Related powers of series C(x) are given as follows.

C(x)^2 = 1 + 2*x^2/2! + 32*x^4/4! + 1376*x^6/6! + 114176*x^8/8! + 15519488*x^10/10! + 3132551168*x^12/12! + 879422726144*x^14/14! + 327670676455424*x^16/16! + 156439068819587072*x^18/18! +...+ A281183(n)*x^(2*n)/(2*n)! +...

where C(x)^2 = 1 + S(x)^2.

C(x)^3 = 1 + 3*x^2/2! + 57*x^4/4! + 2739*x^6/6! + 246801*x^8/8! + 35822307*x^10/10! + 7636142793*x^12/12! + 2246286827091*x^14/14! + 871869519033249*x^16/16! + 431649452286233283*x^18/18! +...+ A281184(n)*x^(2*n)/(2*n)! +...

where C(x)^3 = d/dx log( C(x) + S(x) ).

C(x)^4 = 1 + 4*x^2/2! + 88*x^4/4! + 4672*x^6/6! + 454144*x^8/8! + 70084096*x^10/10! + 15728822272*x^12/12! + 4836914249728*x^14/14! + 1952137912385536*x^16/16! + 1000749157519458304*x^18/18! +...

where C(x)^4 = d/dx S(x).

From Peter Bala, Feb 18 2026: (Start)

Reducing a(n) mod 7 produces the sequence [1, 4, 4, 3, 5, 5, 2, 1, 1, 6, 3, 3, 4, 2, 2, 5, 6, 6, 1, 4, 4, 3, 5, 5, 2, 1, 1, 6, 3, 3, 4, 2, 2, 5, 6, 6, 1, 4, 4, ...], which appears to be a purely periodic sequence with period equal to 18, a divisor of phi(7)^2 = 36. The period can be arranged into 6 = phi(7) columns each of length 3 = phi(7)/2 as follows.

1, 4, 4, 3, 5, 5,

2, 1, 1, 6, 3, 3,

4, 2, 2, 5, 6, 6,

Notice that, working modulo 7, the first column consists of the powers of 2 modulo 7 and the other columns are all multiples of the first column. This 2-dimensional structure of the period appears to hold more generally when reducing the sequence {a(n)} modulo a fixed positive integer k. (End)

MAPLE

f := proc(n, x) option remember; if n = 1 then 1 else simplify( diff((1 + x^2)^2 * f(n-1, x), x) ) end if end proc:

seq( eval(f(2*n-1, x), x = 0), n = 1..20 ); # Peter Bala, Dec 22 2025

MATHEMATICA

nMax = 30; m = maxExponent = 2*nMax; a[n_] := Module[{S = x, C = 1}, For[i = 1, i <= n, i++, S = Integrate[C^4 + x*O[x]^m // Normal, x] + O[x]^m // Normal; C = 1 + Integrate[S*C^3 + O[x]^m // Normal, x]] + O[x]^m // Normal; (2*n - 1)!*Coefficient[S, x, 2*n - 1]]; Table[an = a[n]; Print[ "a(", n, ") = ", an]; an, {n, 1, nMax}] (* Jean-François Alcover, Jan 20 2017, adapted from first PARI program *)

nmax = 20; Table[(CoefficientList[InverseSeries[Series[(x/(1 + x^2) + ArcTan[x])/2, {x, 0, 2*nmax - 1}], x], x] * Range[0, 2*nmax - 1]!)[[2*n]], {n, 1, nmax}] (* Vaclav Kotesovec, Sep 02 2017 *)

PROG

(PARI) {a(n) = my(S=x, C=1); for(i=1, n, S = intformal( C^4 +x*O(x^(2*n))); C = 1 + intformal( S*C^3 ) ); (2*n-1)!*polcoeff(S, 2*n-1)}

for(n=1, 30, print1(a(n), ", "))

(PARI) /* S(x) = Series_Reversion( Integral 1/(1 + x^2)^2 dx ) */

{a(n) = my(S=x); S = serreverse( intformal( 1/(1 + x^2 +x*O(x^(2*n)))^2)); (2*n-1)!*polcoeff(S, 2*n-1)}

for(n=1, 30, print1(a(n), ", "))

CROSSREFS

Column k=0 of A394012.

Cf. A281181 (C), A281182 (C+S), A281183 (C^2), A281184 (C^3), A281427, A281428, A281435, A281436.

Sequence in context: A154136 A367252 A012830 * A296465 A012946 A013101

Adjacent sequences: A281177 A281178 A281179 * A281181 A281182 A281183

KEYWORD