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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). 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
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.
LINKS
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,...
EXAMPLE
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;
PROG
(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))
CROSSREFS
A369993 (denominator).
Sequence in context: A329544 A095006 A159587 * A124732 A167552 A094787
KEYWORD
sign,tabf,frac
AUTHOR
Roland Miyamoto, Mar 01 2024
STATUS
approved

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Last modified June 12 10:07 EDT 2024. Contains 373329 sequences. (Running on oeis4.)