A352798 a(n) = 1/(cf[0;n,n,n,...,n] - cf[0;n,n,...,n]) where the first continued fraction has n+1 terms and the second has n terms.

1, -10, 330, -21960, 2551640, -461930274, 120572270007, -42930583856160, 20008932768992430, -11825788272679695050, 8643081649999714376976, -7654102744143874729100040, 8076084821027629176909996013, -10010473694454865001226770534530, 14402393216408406872433735669683370

Offset: 1

Author

Sebastian F. Orellana, Apr 03 2022

Keywords

• sign
• easy

Examples

			a(2) = -10 because the two continued fractions are cf[0;2,2] = 0 + 1/(2 + 1/2) = 2/5 and cf[0;2] = 0 + 1/2 = 1/2 and the reciprocal of their difference is 1/(2/5 - 1/2) = -10.
a(3) = 330 because the two continued fractions are cf[0;3,3,3] = 0 + 1/(3 + 1/(3 + 1/3)) = 10/33 and cf[0;3,3] = 0 + 1/(3 + 1/3) = 3/10, and 1/(10/33 - 3/10) = 330.
		

Crossrefs

Cf. A084844, A084845.

Programs

Maple

a:= n-> (f-> -(-1)^n*f(n,n)*f(n+1,n))(combinat[fibonacci]):
seq(a(n), n=1..15);  # Alois P. Heinz, Jul 06 2022

PARI

a(n) = (-1)^(n+1) * vecprod(Vec(lift(Mod('x,'x^2-n*'x-1)^(n+1)))); \\ Kevin Ryde, Apr 18 2022

Python

from sympy.ntheory.continued_fraction import continued_fraction_reduce
def A352798(n): return int(1/(continued_fraction_reduce([0]+[n]*n)-continued_fraction_reduce([0]+[n]*(n-1)))) # Chai Wah Wu, Jul 06 2022

Formula

a(n) = A084844(n)*A084845(n)*(-1)^(n+1).

Extensions

More terms from Kevin Ryde, Apr 18 2022