A369992 - OEIS

%I #18 Apr 21 2024 19:15:08

%S 1,-1,1,-3,2,5,-4,-35,28,70,-100,35,3575,-5720,-6292,19240,-14300,

%T 3520,-13856700,22170720,24387792,-74574240,217088300,-401631120,

%U -382444920,2019752592,-1656568485,-1470440400,3671101720,-2832601200,1025395800,-147804800

%N Irregular triangle read by rows: T(n,k) = (2^floor(n/2)+k)-th numerator coefficient of the polynomial q_n used to parametrize the canonical stribolic iterates h_n (of order 1), for n=0,1,2,... and 0 <= k <= A000045(n+1) - 2^floor(n/2).

%C The n-th row of the triangle contains 1 + A000045(n+1) - 2^floor(n/2) integers c_{2^floor(n/2)},...,c_{A000045(n+1)} forming a polynomial q_n = (n mod 2) + Sum_{i} c_i*X^i / A369993(n) that is related to A369990 and A369991 as follows: q_n = h_n ° ... ° h_1 (function composition), that is, h_n maps q_{n-1}(t) to q_n(t) for 0 <= t <= 1, and h_n has Integral_{x=0..1} h_n(x) dx = A369990(n)/A369991(n).

%C The gcd of each row in the triangle equals 1.

%C All previous statements are proved in the arXiv article, see link below.

%C Observation: In each of the 25 rows computed so far, there are no zeros and at most two consecutive entries of the same sign.

%H Roland Miyamoto, <a href="/A369992/b369992.txt">Table of n, a(n) for n = 0..1228</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 The polynomials q_n = (n mod 2) + Sum_{k>=0} T(n,k)*X^(2^floor(n/2)+k) / A369993(n) are determined by the equations 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,...

%F Sum_k T(n,k) = (-1)^n * A369993(n) for n=0,1,2,...

%e q_5 = 1 + (-35*X^4 + 28*X^5 + 70*X^6 - 100*X^7 + 35*X^8) / 2 gives rise to row 5 (counting from 0) of the triangle (rows 0 to 7 are given):

%e           1;

%e          -1;

%e           1;

%e          -3,        2;

%e           5,       -4;

%e         -35,       28,       70,      -100,        35;

%e        3575,    -5720,    -6292,     19240,    -14300,       3520;

%e   -13856700, 22170720, 24387792, -74574240, 217088300, -401631120, -382444920, 2019752592, -1656568485, -1470440400, 3671101720, -2832601200, 1025395800, -147804800;

%o (Python)

%o from functools import cache, reduce; from sympy.abc import x; from sympy import lcm, fibonacci

%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 def row(n): qn = q(n); k0 = 1<<(n>>1); k1 = 1+fibonacci(n+1); dn = reduce(lcm,(denom(qn.coeff(x,k)) for k in range(k0,k1))); return [qn.coeff(x,k)*dn for k in range(k0,k1)]

%o for n in range(15): print(row(n))

%Y A369993 (denominator).

%Y Cf. A369990, A369991, A369988.

%K sign,tabf,frac

%O 0,4

%A _Roland Miyamoto_, Mar 01 2024