%I #33 Apr 20 2024 09:03:44

%S 1,2,3,10,7,572,89148,177918244665,11711158115225119429452,

%T 8990773234863161759100003096510729982749072312,

%U 140048278006628885452600904137492554179859017924910241263151850844470542993943699969398879

%N Denominator of canonical iterated stribolic area Integral_{t=0..1} h_n(t) dt (of order 1).

%C a(n) = denominator of Integral_{t=0..1} h_n(t) dt, where h_0 = 1, h_1 = T(h_0), h_2 = T(h_1), ...:[0,1]->[0,1], the operator T is given by T(g)(x) := Integral_{y=x..1} g^*(y) dy / Integral_{y=0..1} g(y)dy and g^*(y) := sup g^{-1}[y,1] (pseudo-inverse).

%C Geometrically speaking, T rotates by 90 degrees before integrating, which is why we call h_0, h_1, h_2, ... the canonical stribolic iterates (from Greek stribo=turn/twist).

%C Alternatively, a(n) can be calculated from the polynomial q_n := h_n ° ... ° h_1. Cf. alternative formula below.

%C The sequence (a(n)/A369991(n)) is strictly decreasing and converges to the stribolic constant kappa=A369988.

%H Roland Miyamoto, <a href="/A369991/b369991.txt">Table of n, a(n) for n = 0..14</a>

%H Roland Miyamoto, <a href="https://timebrain.org/Levine/B369991.txt">Table of n, a(n) for n = 0..23</a>

%H Roland Miyamoto, <a href="https://arxiv.org/abs/2402.06618">Polynomial parametrisation of the canonical iterates to the solution of -gamma*g' = g^{-1}</a>, arXiv:2402.06618 [math.CO], 2024.

%F a(0)=1, a(n) is the denominator of kappa_n := Integral_{t=0..1} h_n(t) dt where h_1(x):=1-x and h_{n+1}(x) := Integral_{t=x..1} h_n^*(t) dt / kappa_n for n=1,2,...; here, h_n^* denotes the compositional inverse of h_n.

%F Alternatively, the rational sequence (kappa_n) := (A369990(n)/a(n)) and the two polynomial sequences (q_n), (Q_n) together are determined by the following equations for n=1,2,...: kappa_0=1, q_0=X, q_1=1-X, Q_n(0)=0, Q_n' = q_n'*q_{n-1}, kappa_n = (-1)^n * Q_n(1), q_{n+1} = (n+1) mod 2 - Q_n / kappa_n.

%e h_2(x) = (1-x)^2, h_2^*(x) = 1 - sqrt(x) = - h_3'(x)/3, h_3(x) = 1 - 3x + 2x^(3/2), hence Integral_{t=0..1} h_2(t) dt = 1/3 and Integral_{t=0..1} h_3(t) dt = 3/10. Therefore a(2)=3 and a(3)=10.

%o (Python)

%o from functools import cache; from sympy.abc import x

%o @cache

%o def kappa(n): return (1-(n%2)*2) * Q(n).subs(x,1) if n else 1

%o @cache

%o def Q(n): return (q(n).diff() * q(n-1)).integrate()

%o @cache

%o def q(n): return (1-x if n==1 else n%2-Q(n-1)/kappa(n-1)) if n else x

%o def denom(c): return c.denominator() if c%1 else 1

%o print([denom(kappa(n)) for n in range(15)])

%Y Cf. A369988 (decimal expansion of limit), A369990 (numerator).

%Y Cf. A369992, A369993.

%K nonn,frac

%O 0,2

%A _Roland Miyamoto_, Feb 08 2024