|
%I #18 Jan 26 2022 15:00:49
%S 0,1,2,6,15,11,10,45,99,79,65,312,675,545,442,2142,4623,3739,3026,
%T 14685,31683,25631,20737,100656,217155,175681,142130,689910,1488399,
%U 1204139,974170,4728717,10201635,8253295,6677057,32411112,69923043,56568929,45765226,222149070,479259663,387729211,313679522
%N a(n) = F(n)*F(n+1) mod L(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.
%H Robert Israel, <a href="/A348592/b348592.txt">Table of n, a(n) for n = 0..4759</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (0,-1,1,5,3,2,1).
%F For n >= 1, a(n) = (A070352(n+2)*A000032(n+2) + 3*(-1)^n)/5.
%F a(n) + 2*a(n + 1) + 3*a(n + 2) + 5*a(n + 3) + a(n + 4) - a(n + 5) - a(n + 7) = 0 for n >= 1.
%F G.f.: -3 + 3/(5*(1+x)) + (3+x)/(2*(1-x-x^2)) + (9-4*x+6*x^2-x^3)/(10*(1+3*x^2+x^4)).
%e a(5) = F(5)*F(6) mod L(7) = 5*8 mod 29 = 11.
%p F:= combinat:-fibonacci:
%p L:= n -> F(n-1)+F(n+1):
%p seq(F(n)*F(n+1) mod L(n+2), n=0..20);
%t a[n_] := Mod[Fibonacci[n] * Fibonacci[n + 1], LucasL[n + 2]]; Array[a, 50, 0] (* _Amiram Eldar_, Jan 26 2022 *)
%Y Cf. A000032, A000045, A070352, A333599, A347861, A348591.
%K nonn,easy
%O 0,3
%A _J. M. Bergot_ and _Robert Israel_, Jan 25 2022
|
