A369992 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).

1, -1, 1, -3, 2, 5, -4, -35, 28, 70, -100, 35, 3575, -5720, -6292, 19240, -14300, 3520, -13856700, 22170720, 24387792, -74574240, 217088300, -401631120, -382444920, 2019752592, -1656568485, -1470440400, 3671101720, -2832601200, 1025395800, -147804800

Offset: 0

Author

Roland Miyamoto, Mar 01 2024

Keywords

• sign
• tabf
• frac

Comments

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).

The gcd of each row in the triangle equals 1.

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

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

Examples

			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):
          1;
         -1;
          1;
         -3,        2;
          5,       -4;
        -35,       28,       70,      -100,        35;
       3575,    -5720,    -6292,     19240,    -14300,       3520;
  -13856700, 22170720, 24387792, -74574240, 217088300, -401631120, -382444920, 2019752592, -1656568485, -1470440400, 3671101720, -2832601200, 1025395800, -147804800;
		

Links

• Roland Miyamoto, Table of n, a(n) for n = 0..1228
• Roland Miyamoto, Polynomial parametrisation of the canonical iterates to the solution of -gamma*g' = g^{-1}, arXiv:2402.06618 [math.CO], 2024.

Crossrefs

A369993 (denominator).

Cf. A369990, A369991, A369988.

Programs

Python

from functools import cache, reduce; from sympy.abc import x; from sympy import lcm, fibonacci
@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
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)]
for n in range(15): print(row(n))

Formula

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,...

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