OFFSET
0,3
COMMENTS
From Peter Bala, Mar 03 2026: (Start)
Conjectures:
1) If N is not divisble by either 2 or 3 then there exists n_0 such that N divides a(n) for all n >= n_0.
2) If N = 2^a * 3^b * N', where at least one of a, b > 0, then the sequence obtained by reducing a(n) modulo N is eventually periodic and the period divides phi(N), where phi(n) = A000010(n).
Examples are given below. (End)
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..200
FORMULA
E.g.f. C(x) = ( d/dx Series_Reversion( x - x^3/3 ) )^(1/2).
E.g.f. C(x) = ( d/dx Series_Reversion( sin(x) - sin(x)^3/3 ) )^(1/3).
E.g.f. C(x) = ( d/dx Series_Reversion( sinh(x)*(2 + cosh(2*x))/(3*cosh(x)^3) ) )^(1/4).
E.g.f. C(x) = ( d/dx Series_Reversion( x*sqrt(1+x^2)*(3 + 2*x^2)/(3*(1 + x^2)^2) ) )^(1/5).
E.g.f. C(x) = d/dx Series_Reversion( Integral 1/G(x) dx ) where G(x) = e.g.f. of A281181.
E.g.f. C(x) = ( d/dx Series_Reversion( Integral (1 - x^2) dx ) )^(1/2).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral cos(x)^3 dx ) )^(1/3).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral 1/cosh(x)^4 dx ) )^(1/4).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ) )^(1/5).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral G(i*x)^6 dx ) )^(1/6) where G(x) = e.g.f. of A281181.
E.g.f. C(x) and related series S(x) (e.g.f. of A281427) satisfy:
(1.a) C(x)^2 - S(x)^2 = 1.
(1.b) C(x)^2 + S(x)^2 = 1 + Integral 4*C(x)^5*S(x) dx.
Integrals.
(2.a) S(x) = Integral C(x)^5 dx.
(2.b) C(x) = 1 + Integral C(x)^4*S(x) dx.
Exponential.
(3.a) C(x) + S(x) = exp( Integral C(x)^4 dx ).
(3.b) C(x) = cosh( Integral C(x)^4 dx ).
(3.c) S(x) = sinh( Integral C(x)^4 dx ).
Derivatives.
(4.a) S'(x) = C(x)^5.
(4.b) C'(x) = C(x)^4*S(x).
(4.c) (C'(x) + S'(x))/(C(x) + S(x)) = C(x)^4.
(4.d) (C(x)^2 + S(x)^2)' = 4*C(x)^5*S(x).
Explicit Solutions.
(5.a) S(x) = Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ).
(5.b) C(x)^1 = d/dx Series_Reversion( Integral 1/G(x) dx ) where G(x) = e.g.f. of A281181.
(5.c) C(x)^2 = d/dx Series_Reversion( Integral (1 - x^2) dx ).
(5.d) C(x)^3 = d/dx Series_Reversion( Integral cos(x)^3 dx ).
(5.e) C(x)^4 = d/dx Series_Reversion( Integral 1/cosh(x)^4 dx ).
(5.f) C(x)^5 = d/dx Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ).
(5.g) C(x)^6 = d/dx Series_Reversion( Integral G(i*x)^6 dx ) where G(x) = e.g.f. of A281181.
(5.h) C(x)^2 = d/dx Series_Reversion( x - x^3/3 ).
(5.j) C(x)^3 = d/dx Series_Reversion( sin(x) - sin(x)^3/3 ).
(5.j) C(x)^4 = d/dx Series_Reversion( sinh(x)*(2 + cosh(2*x))/(3*cosh(x)^3) ).
(5.k) C(x)^5 = d/dx Series_Reversion( x*sqrt(1+x^2)*(3 + 2*x^2)/(3*(1 + x^2)^2) ).
From Peter Bala, Dec 19 2025: (Start)
The g.f. C(x) is algebraic: (4 - 9*x^2)*C(x)^6 - 3*C(x)^2 - 1 = 0 with C(0) = 1.
C(x)^2 = Sum_{n >= 0} A052502(n)*x^(2*n)/(2*n)!.
C(x)^2 = Sum_{n >= 0} binomial(3*n,n)*x^(2*n)/3^n. (End)
From Peter Bala, Jan 22 2026: (Start)
Define d(n, x) = 1/(1 - x^2)*d/dx( d(n-1, x) ) with d(0, x) = 1/sqrt(1 - x^2). Then a(n) = d(2*n, 0).
Define f(n, x) = d/dx( sec^3(x)*f(n-1, x) ) with f(0, x) = (cos(x))^2. Then a(n) = f(2*n, 0).
More generally, if we define f_k(n, x) = d/dx( sec^3(x)*f(n-1, x) ) with f(0, x) = (sec(x))^(k-3), k arbitrary, then f_k(n, 0) = [x^n] C(x)^k. (End)
From Vaclav Kotesovec, Feb 24 2026: (Start)
Recurrence: 16*a(n) = (288*n^2 - 576*n + 317)*a(n-1) - 9*(2*n - 3)^2*(6*n - 11)*(6*n - 7)*a(n-2).
a(n) ~ sqrt(2*Pi) * 3^(2*n + 1/4) * n^(2*n - 1/4) / (Gamma(1/4) * exp(2*n)). (End)
Let A(n,k) = (2*n)! * [x^(2*n)] C(x)^k. A(0,k) = 1 and A(n,k) = k*(k+4) * A(n-1,k+8) - k*(k+3) * A(n-1,k+6) for n > 0. a(n) = A(n,1). - Seiichi Manyama, Apr 14 2026
EXAMPLE
E.g.f.: C(x) = 1 + x^2/2! + 17*x^4/4! + 865*x^6/6! + 88865*x^8/8! + 15335425*x^10/10! + 3993275825*x^12/12! + 1462392957025*x^14/14! + 716611617346625*x^16/16! + 452780458211706625*x^18/18! + 358439197464543820625*x^20/20! +...
such that
(1) C(x) = cosh( Integral C(x)^4 dx ),
(2) C(x)^2 - S(x)^2 = 1, and
(3) C(x) = 1 + Integral C(x)^4*S(x) dx,
where S(x) is described by A281427 and begins:
S(x) = x + 5*x^3/3! + 145*x^5/5! + 10325*x^7/7! + 1357825*x^9/9! + 284963525*x^11/11! + 87274812625*x^13/13! + 36716097543125*x^15/15! + 20309401097610625*x^17/17! + 14290053364475013125*x^19/19! +...
From Peter Bala, Mar 03 2026: (Start)
Reducing a(n) mod N, where N = 5*7*11 = 385, produces the sequence [1, 1, 17, 95, 315, 105, 0, 0, 0, 0, 0, 0, 0, 0, ...]. It appears that 385 divides a(n) for n >= 6.
Reducing a(n) mod 3^3 produces the sequence [1, 1, 17, 1, 8, 19, 17, 1, 26, 19, 17, 10, 26, 19, 8, 10, 26, 1, 8, 10, 17, 1, 8, 19, 17, 1, 26, 19, 17, 10, 26, 19, 8, 10, 26, 1, 8, 10, 17, 1, 8, 19, 17, 1, 26, ...], which appears to be an eventually periodic sequence with a period of 18 = phi(3^3) beginning at n = 7. (End)
MAPLE
f := proc (n, x) option remember; if n = 0 then cos(x)^2 else simplify( diff( sec(x)^3 * f(n-1, x), x) ) end if end proc:
seq( eval(f(2*n, x), x = 0), n = 0..20 ); # Peter Bala, Jan 22 2026
MATHEMATICA
a[n_] := Module[{S = x, C = 1, C5, SC4}, For[i = 0, i <= n, i++, C5 = C^5 + x*O[x]^(2n) // Normal; S = Integrate[C5 , x]; SC4 = S*C^4+O[x]^(2n) // Normal; C = 1 + Integrate[SC4, x]]; (2n)!*Coefficient[C, x, 2n]]; Array[a, 17, 0] (* Jean-François Alcover, Mar 01 2017, translated from Pari *)
nmax = 20; Table[CoefficientList[Sqrt[D[InverseSeries[Series[x - x^3/3, {x, 0, 2*nmax + 1}], x], x]], x] [[2*n + 1]], {n, 0, nmax}] * (2*Range[0, nmax])! (* Vaclav Kotesovec, Feb 24 2026 *)
PROG
(PARI) {a(n) = my(S=x, C=1); for(i=0, n, S = intformal( C^5 +x*O(x^(2*n))); C = 1 + intformal( S*C^4 ) ); (2*n)!*polcoeff(C, 2*n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) a(n, k=1) = if(n==0, 1, k*(k+4)*a(n-1, k+8)-k*(k+3)*a(n-1, k+6)); \\ Seiichi Manyama, Apr 14 2026
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jan 21 2017
STATUS
approved