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Product of Lucas numbers and inverted Lucas numbers: a(n)=A000032(n)*A075193(n).
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%I #27 Jul 09 2024 07:46:08

%S 2,-3,12,-28,77,-198,522,-1363,3572,-9348,24477,-64078,167762,-439203,

%T 1149852,-3010348,7881197,-20633238,54018522,-141422323,370248452,

%U -969323028,2537720637,-6643838878,17393796002,-45537549123,119218851372,-312119004988,817138163597

%N Product of Lucas numbers and inverted Lucas numbers: a(n)=A000032(n)*A075193(n).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-2,2,1).

%F a(n) = 1 + (-1)^n*A002878(n).

%F From _Michael Somos_, Apr 07 2003: (Start)

%F G.f.: (2+x+2x^2)/((1+3x+x^2)(1-x)).

%F a(n) = -3a(n-1) - a(n-2)+5 = -2a(n-1) + 2a(n-2) + a(n-3) = a(-1-n). (End)

%F Sum_{n>=0} 1/a(n) = sqrt(5)/10. - _Amiram Eldar_, Jan 15 2022

%F a(n) = (-1)^n*A215602(n). - _R. J. Mathar_, Jul 09 2024

%F a(n) - a(n-1) = (-1)^n* A054888(n), n>0. - _R. J. Mathar_, Jul 09 2024

%t CoefficientList[Series[(2 + x + 2x^2)/(1 + 2x - 2x^2 - x^3), {x, 0, 30}], x]

%t LinearRecurrence[{-2,2,1},{2,-3,12},30] (* _Harvey P. Dale_, Jun 30 2022 *)

%o (PARI) a(n)=1+(-1)^n*(fibonacci(2*n)+fibonacci(2*n+2))

%Y Cf. A000032, A002878, A075193.

%K easy,sign

%O 0,1

%A Mario Catalani (mario.catalani(AT)unito.it), Sep 11 2002