OFFSET
0,3
COMMENTS
From Paul Curtz, Jan 20 2017: (Start)
a(n) mod 10 = periodic sequence of length 8: repeat [1, 1, 3, 3, 9, 9, 7, 7] = duplicated A001148(n).
a(n) mod 9 = 1, followed by period 3: repeat [1, 4, 7]. See A100402. See also A281280, A281182, A281183, A281184 (1, followed by 3's).
a(n+p) - a(n) is a multiple of 12. (End)
More generally, we conjecture that 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 16 2026
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..216 (terms 0..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. C(x) = d/dx Series_Reversion( Integral sqrt(1 - x^2) dx ).
E.g.f. C(x) = d/dx Series_Reversion( ( x*sqrt(1 - x^2) + asin(x) )/2 ).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral cos(x)^2 dx ) )^(1/2).
E.g.f. C(x) = ( d/dx Series_Reversion( (2*x + sin(2*x))/4 ) )^(1/2).
E.g.f. C(x) = ( d/dx Series_Reversion( Integral 1/cosh(x)^3 dx ) )^(1/3).
E.g.f. C(x) = ( d/dx Series_Reversion( ( sinh(x)/cosh(x)^2 + atan(sinh(x)) )/2 ) )^(1/3).
E.g.f. C(x) and related series S(x) (e.g.f. of A281180) 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 ).
(5.f) C(x)^4 = d/dx Series_Reversion( Integral 1/(1 + x^2)^2 dx ).
(5.g) C(x)^5 = d/dx Series_Reversion( Integral C(i*x)^5 dx ).
From Peter Bala, Dec 14 2025: (Start)
Let T(x) = S(x)/C(x) = x + x^3/3! + 13*x^5/6! + 493*x^7/7! + 37369*x^9/9! + 4732249*x^11/11! + 901188997*x^13/13! + .... Then d/dx(T(x)) = C(x).
T(x)^2 + 1/C(x)^2 = 1. Cf. tanh(x)^2 + 1/cosh(x)^2 = 1.
d/dx( Series_Reversion(T(x)) ) = sqrt(1 - x^2). Cf. d/dx( Series_Reversion(tanh(x)) ) = 1/(1 - x^2).
T(x) = Series_Reversion( (x*sqrt(1 - x^2) + arcsin(x))/2 ). (End)
From Peter Bala, Jan 26 2026: (Start)
C(x) = sec( Series_Reversion(x/2 + sin(2*x)/4) ).
For n >= 1, define f(n, x) = d/dx( 1/sqrt(1 - x^2) * f(n-1, x) ) with f(0, x) = 1. Then
a(n) = f(2*n, 0) (apply Dominici, Theorem 4.1, to C(x) = d/dx Series_Reversion( Integral sqrt(1 - x^2) dx )).
More generally, for n >= 1, define f_k(n, x) = d/dx( 1/sqrt(1 - x^2) * f(n-1, x) ) with f_k(0, x) = 1/(1 - x^2)^(k/2). Then f_k(2*n, 0) gives the coefficients of C(x)^(k+1).
For n >= 1, define g(n, x) = (sec(x))^2 * d/dx(g(n-1, x)) with g(0, x) = sec(x). Then a(n) = g(2*n, 0). (End)
Let A(n,k) = (2*n)! * [x^(2*n)] C(x)^k. A(0,k) = 1 and A(n,k) = k*(k+3) * A(n-1,k+6) - k*(k+2) * A(n-1,k+4) for n > 0. a(n) = A(n,1). - Seiichi Manyama, Apr 14 2026
EXAMPLE
E.g.f.: 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! +...
such that
(1) C(x) = cosh( Integral C(x)^3 dx ),
(2) C(x)^2 - S(x)^2 = 1, and
(3) C(x) = 1 + Integral C(x)^3*S(x) dx,
where S(x) begins:
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! +...+ A281180(n)*x^(2*n-1)/(2*n-1)! +...
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 +...
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) ).
Also, C(x)^3 = d/dx Series_Reversion( Integral 1/cosh(x)^3 dx ).
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! +...+ A281180(n+1)*x^(2*n)/(2*n)! +...
where C(x)^4 = d/dx S(x).
From Peter Bala, Feb 16 2026: (Start)
Reducing a(n) mod 13 produces the sequence [1, 1, 0, 12, 7, 2, 7, 7, 0, 6, 10, 1, ...], which appears to be a purely periodic sequence with period equal to 72, a divisor of phi(13)^2 = 144. The period can be arranged into 6 = phi(13)/2 columns each of length 12 = phi(13).
1, 1, 0, 12, 7, 2,
7, 7, 0, 6, 10, 1,
10, 10, 0, 3, 5, 7,
5, 5, 0, 8, 9, 10,
9, 9, 0, 4, 11, 5,
11, 11, 0, 2, 12, 9,
12, 12, 0, 1, 6, 11,
6, 6, 0, 7, 3, 12,
3, 3, 0, 10, 8, 6,
8, 8, 0, 5, 4, 3,
4, 4, 0, 9, 2, 8,
2, 2, 0, 11, 1, 4,
Notice that, working modulo 13, the first column consists of the powers of 7 modulo 13 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 = 0 then 1 else simplify( diff(1/sqrt(1-x^2)*f(n-1, x), x) ) end if; end proc:
seq( eval(f(2*n, x), x = 0), n = 0..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)!*Coefficient[C, x, 2*n]]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, nMax}] (* Jean-François Alcover, Jan 20 2017, adapted from PARI *)
nmax = 20; Table[(CoefficientList[Sqrt[D[InverseSeries[Series[(2*x + Sin[2*x])/4, {x, 0, 2*nmax - 1}], x], x]], x] * Range[0, 2*nmax - 2]!)[[2*n - 1]], {n, 1, nmax}] (* Vaclav Kotesovec, Sep 02 2017 *)
PROG
(PARI) {a(n) = my(S=x, C=1); for(i=0, n, S = intformal( C^4 +x*O(x^(2*n))); C = 1 + intformal( S*C^3 ) ); (2*n)!*polcoeff(C, 2*n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) a(n, k=1) = if(n==0, 1, k*(k+3)*a(n-1, k+6)-k*(k+2)*a(n-1, k+4)); \\ Seiichi Manyama, Apr 14 2026
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jan 16 2017
EXTENSIONS
Name simplified by Paul D. Hanna, Jan 22 2017
STATUS
approved