OFFSET
0,2
COMMENTS
Conjecture: the sequence {a(n)} taken 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
Paul D. Hanna, Table of n, a(n) for n = 0..100
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)^2 = d/dx Series_Reversion( Integral cos(x)^2 dx ).
E.g.f. C(x)^2 = d/dx Series_Reversion( (2*x + sin(2*x))/4 ).
E.g.f. C(x)^2 where related series S(x) and C(x) 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)^4 - S(x)^4)' = 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 ).
From Peter Bala, Dec 22 2025: (Start)
For n >= 1, define f(n, x) = d/dx( sec(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)^2 = d/dx Series_Reversion( Integral cos(x)^2 dx )).
For n >= 1, define g(n, x) = (1 + x^2) * d/dx( (1 + x^2)*g(n-1, x) ) with g(0, x) = 1. Then it appears that a(n) = g(2*n, 0). (End)
For n >= 1, define h(n, x) = d/dx( 1/sqrt(1 - x^2)*h(n-1, x) ) with h(0, x) = 1/sqrt(1 - x^2). Then a(n) = h(2*n, 0). - Peter Bala, Jan 09 2026
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,2). - Seiichi Manyama, Apr 14 2026
EXAMPLE
E.g.f.: 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! +...
such that C(x)^2 = 1 + S(x)^2 and the series for C(x) and S(x) begin:
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! +...+ A281181(n)*x^(2*n)/(2*n)! +...
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! +...+ A281180(n)*x^(2*n-1)/(2*n-1)! +...
From Peter Bala, Feb 16 2026: (Start)
Reducing a(n) mod 11 produces the sequence [1, 2, 10, 1, 7, 6, 10, 6, 5, 2, 8, 6, 8, 3, 10, 7, 8, 7, 4, 6, 2, 7, 2, 9, 8, 10, 2, 10, 1, 7, 6, ...], which, starting at a(1) = 2, appears to be a periodic sequence with period equal to 25, a divisor of phi(11)^2 = 100. The period can be arranged into 5 = phi(11)/2 columns each of length 5 as follows.
2, 10, 1, 7, 6,
10, 6, 5, 2, 8
6, 8, 3, 10, 7,
8, 7, 4, 6, 2,
7, 2, 9, 8, 10,
Notice that, working modulo 11, the first column consists of the numbers 2*5^n modulo 11 for 0 <= n <= 4, 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( sec(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
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)!*polcoeff(C^2, 2*n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* From C(x)^2 = d/dx Series_Reversion( Integral cos(x)^2 dx ) */
{a(n) = my(C2=x); C2 = deriv( serreverse( intformal( cos(x +x*O(x^(2*n)))^2 ))); (2*n)!*polcoeff(C2, 2*n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) a(n, k=2) = 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
STATUS
approved