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a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.
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%I #42 Jan 27 2022 09:43:05

%S 0,1,0,3,5,3,18,3,52,3,141,3,374,3,984,3,2581,3,6762,3,17708,3,46365,

%T 3,121390,3,317808,3,832037,3,2178306,3,5702884,3,14930349,3,39088166,

%U 3,102334152,3,267914293,3,701408730,3,1836311900,3,4807526973,3,12586269022,3,32951280096,3,86267571269,3

%N a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

%H Robert Israel, <a href="/A348591/b348591.txt">Table of n, a(n) for n = 0..4761</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-1,3,3,-1,-1).

%F a(n) = 3 if n >= 3 is odd.

%F a(n) = A000045(n+2)-3 if n >= 2 is even.

%F a(n) + a(n+1) - 3*a(n+2) - 3*a(n+3) + a(n+4) + a(n+5) = 0 for n >= 2.

%F G.f.: -x*(2*x^5-5*x^3-x-1)/((x+1)*(x^2+x-1)*(x^2-x-1)). - _Alois P. Heinz_, Jan 26 2022

%e a(5) = L(5)*L(6) mod F(7) = 11*18 mod 13 = 3.

%p F:= combinat:-fibonacci:

%p L:= n -> F(n-1)+F(n+1):

%p map(n -> L(n)*L(n+1) mod F(n+2), [$0..30]);

%t a[n_] := Mod[LucasL[n] * LucasL[n + 1], Fibonacci[n + 2]]; Array[a, 50, 0] (* _Amiram Eldar_, Jan 26 2022 *)

%o (Python)

%o from gmpy2 import fib, lucas2

%o def A348591(n): return (lambda x,y:int(x[0]*x[1] % y))(lucas2(n+1),fib(n+2)) # _Chai Wah Wu_, Jan 26 2022

%Y Cf. A000032, A000045, A215602, A333599, A347861, A348592.

%K nonn,easy

%O 0,4

%A _J. M. Bergot_ and _Robert Israel_, Jan 25 2022