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(7*a(n))^2 - 53*b(n)^2 = -4 with b(n)=A097838(n) give gives all positive solutions of this Pell equation.

a(n) = 51*a(n-1) - a(n-2), ; a(0)=1, a(1)=52. - Philippe Deléham, Nov 18 2008

b(n) = (1/2)*( (-1)^n - 1 )*F(n,7) + (1/7)*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).

(MAGMAMagma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019

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Cf. A097783, A097834, A097836, A097840.

b(n) = 1/2*( (-1)^n - 1 )*F(n,7) + 1/7*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1). (End)

Exp( Sum_{n >= 1} 14*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*x^n.

Exp( Sum_{n >= 1} (-14)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*(-x)^n. (End)

H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

From Peter Bala, Aug 26 2022: (Start)

a(n) = (2/7)*(7/2 o 7/2 o ... o 7/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.

The aerated sequence (b(n))n>=1 = [1, 0, 52, 0, 2651, 0, 135149, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 51*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -49, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.

b(n) = 1/2*( (-1)^n - 1 )*F(n,7) + 1/7*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1). (End)

Cf. A097783, A097834, A097836.

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<a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (51, -1).

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