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A369993 Reciprocal of content of the polynomial q_n used to parametrize the canonical stribolic iterates h_n (of order 1).

%I A369993 #19 Jun 27 2025 15:08:05
%S A369993 1,1,1,1,1,2,23,24941,1307261674,62079371576837874658793,
%T A369993 67775687882486213674661973555079371183525163,
%U A369993 39058362193701767718721504578116138158143785410766642680982462728116470023287868511995843
%N A369993 Reciprocal of content of the polynomial q_n used to parametrize the canonical stribolic iterates h_n (of order 1).
%C A369993 1/a(n) is the content of the polynomial q_n, whose (non-constant) numerator coefficients are given by A369992, that is, a(n)*q_n in Z[X] is primitive. (Proof in arXiv article, see link below.)
%H A369993 Roland Miyamoto, <a href="/A369993/b369993.txt">Table of n, a(n) for n = 0..15</a>
%H A369993 Roland Miyamoto, <a href="https://timebrain.org/Levine/B369993.txt">Table of n, a(n) for n = 0..24</a>
%H A369993 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 A369993 1/a(n) = content of the polynomial q_n in Q[X] determined by the identities q_0 = X, q_1 = 1 - X, q_n(0) = n mod 2 and (A369990(n) / A369991(n)) * q_{n+1}' = -q_n' * q_{n-1} for n=1,2,...
%e A369993 q_5 = 1 + ( -35*X^4 + 28*X^5 + 70*X^6 - 100*X^7 + 35*X^8 ) / 2 and q_6 = ( 3575*X^8 - 5720*X^9 - 6292*X^10 + 19240*X^11 - 14300*X^12 + 3520*X^13 ) / 23.
%e A369993 Therefore, a(5)=2 and a(6)=23.
%o A369993 (Python)
%o A369993 from functools import cache, reduce; from sympy.abc import x; from sympy import lcm, fibonacci
%o A369993 @cache
%o A369993 def kappa(n): return (1-(n%2)*2) * Q(n).subs(x,1) if n else 1
%o A369993 @cache
%o A369993 def Q(n): return (q(n).diff() * q(n-1)).integrate()
%o A369993 @cache
%o A369993 def q(n): return (1-x if n==1 else n%2-Q(n-1)/kappa(n-1)) if n else x
%o A369993 def denom(c): return c.denominator if c%1 else 1
%o A369993 def A369993(n): return reduce(lcm,(denom(q(n).coeff(x,k)) for k in range(1<<(n>>1),1+fibonacci(n+1))))
%o A369993 print([A369993(n) for n in range(15)])
%Y A369993 Cf. A369992 (triangle of numerators).
%Y A369993 Cf. A369990, A369991, A369988.
%K A369993 nonn,frac
%O A369993 0,6
%A A369993 _Roland Miyamoto_, Mar 01 2024