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A369993
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Reciprocal of content of the polynomial q_n used to parametrize the canonical stribolic iterates h_n (of order 1).
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5
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1, 1, 1, 1, 1, 2, 23, 24941, 1307261674, 62079371576837874658793, 67775687882486213674661973555079371183525163, 39058362193701767718721504578116138158143785410766642680982462728116470023287868511995843
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OFFSET
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0,6
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COMMENTS
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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.)
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LINKS
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FORMULA
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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,...
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EXAMPLE
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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.
Therefore, a(5)=2 and a(6)=23.
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PROG
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(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 A369993(n): return reduce(lcm, (denom(q(n).coeff(x, k)) for k in range(1<<(n>>1), 1+fibonacci(n+1))))
print([A369993(n) for n in range(15)])
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CROSSREFS
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Cf. A369992 (triangle of numerators).
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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