O.g.f.: (2 - 2*x)/(1 - 2*x - x^2). -
Simon Plouffe in his 1992 dissertation
a(n) = (1 + sqrt(2))^n + (1 - sqrt(2))^n. - Mario Catalani (mario.catalani(AT)unito.it), Mar 17 2003
Given F(n) =
A000129(n), the Pell numbers, and L(n) = a(n), then:
L(n+m) + (-1)^m*L(n-m) = L(n)*L(m).
L(n+m) - (-1)^m*L(n-m) = 8*F(n)*F(m).
L(n+m+k) + (-1)^k*L(n+m-k) + (-1)^m*(L(n-m+k) + (-1)^k*L(n-m-k)) = L(n)*L(m)*L(k).
L(n+m+k) - (-1)^k*L(n+m-k) + (-1)^m*(L(n-m+k) - (-1)^k*L(n-m-k)) = 8*F(n)*L(m)*F(k).
L(n+m+k) + (-1)^k*L(n+m-k) - (-1)^m*(L(n-m+k) + (-1)^k*L(n-m-k)) = 8*F(n)*F(m)*L(k).
L(n+m+k) - (-1)^k*L(n+m-k) - (-1)^m*(L(n-m+k) - (-1)^k*L(n-m-k)) = 8*L(n)*F(m)*F(k).
(End)
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(2*k - 1)/(x*(2*k + 1) - 1/G(k + 1))); (continued fraction). -
Sergei N. Gladkovskii, Jun 19 2013
a(n) = [x^n] ( (1 + 2*x + sqrt(1 + 4*x + 8*x^2))/2 )^n for n >= 1. -
Peter Bala, Jun 23 2015
a(n)^2 - a(n + 1) * a(n - 1) = (-1)^(n) * 8.
a(n)^2 - a(n + r) * a(n - r) = (-1)^(n - r - 1) * 8 *
A000129(r)^2.
a(m) * a(n + 1) - a(m + 1) * a(n) = (-1)^(n - 1) * 8 *
A000129(m - n).
(End)
a(n) = (-1)^n * (a(n)^3 - a(3*n))/3. -
Greg Dresden, Jun 16 2021
a(n) = (a(n+2) + a(n-2))/6 for n >= 2. -
Greg Dresden, Jun 23 2021
a(3n+2)/a(3n-1) = [14, ..., 14, -3] with (n+1) 14's.
a(3n+3)/a( 3n ) = [14, ..., 14, 7] with n 14's.
a(3n+4)/a(3n+1) = [14, ..., 14, 17] with n 14's. (End)
a(n) = trace([2, 1; 1, 0]^n) for n >= 1.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all positive integers n and k and all primes p.
a(3^n) ==
A271222(n) (mod 3^n). (End)
Sum_{n>=1} arctan(2/a(n))*arctan(2/a(n+1)) = Pi^2/32 (
A244854) (Ohtsuka, 2019). -
Amiram Eldar, Feb 11 2024
The following series telescope (Cf.
A000032):
For k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)
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