A369991 Denominator of canonical iterated stribolic area Integral_{t=0..1} h_n(t) dt (of order 1).
1, 2, 3, 10, 7, 572, 89148, 177918244665, 11711158115225119429452, 8990773234863161759100003096510729982749072312, 140048278006628885452600904137492554179859017924910241263151850844470542993943699969398879
Offset: 0
Examples
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.
Links
- Roland Miyamoto, Table of n, a(n) for n = 0..14
- Roland Miyamoto, Table of n, a(n) for n = 0..23
- Roland Miyamoto, Polynomial parametrisation of the canonical iterates to the solution of -gamma*g' = g^{-1}, arXiv:2402.06618 [math.CO], 2024.
Programs
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Python
from functools import cache; from sympy.abc import x @cache def kappa(n): return (1-(n%2)*2) * Q(n).subs(x,1) if n else 1 @cache def Q(n): return (q(n).diff() * q(n-1)).integrate() @cache def q(n): return (1-x if n==1 else n%2-Q(n-1)/kappa(n-1)) if n else x def denom(c): return c.denominator if c%1 else 1 print([denom(kappa(n)) for n in range(15)])
Formula
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.
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.
Comments