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a(n) = J(n+1) mod J(n), J(n)=A001045(n).
1

%I #12 Oct 04 2018 18:11:12

%S 1,0,0,2,1,10,1,42,1,170,1,682,1,2730,1,10922,1,43690,1,174762,1,

%T 699050,1,2796202,1,11184810,1,44739242,1,178956970,1,715827882,1,

%U 2863311530,1,11453246122,1,45812984490,1,183251937962,1,733007751850,1,2932031007402

%N a(n) = J(n+1) mod J(n), J(n)=A001045(n).

%H Colin Barker, <a href="/A112691/b112691.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,5,0,-4).

%F G.f.: x*(1-5*x^2+2*x^3+5*x^4-4*x^6) / (1-5*x^2+4*x^4).

%F a(2*n) = 1 - C(1, n) + C(0, n); a(2*n+1) = 2*A002450(n).

%F From _Colin Barker_, Apr 21 2017: (Start)

%F a(n) = (1 - (-2)^n + 5*(-1)^n + 2^n) / 6 for n>2.

%F a(n) = 5*a(n-2) - 4*a(n-4) for n>4.

%F (End)

%t LinearRecurrence[{0,5,0,-4},{1,0,0,2,1,10,1},50] (* _Harvey P. Dale_, Oct 04 2018 *)

%o (PARI) Vec((1 - 5*x^2 + 2*x^3 + 5*x^4 - 4*x^6) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + 2*x)) + O(x^30)) \\ _Colin Barker_, Apr 21 2017

%K easy,nonn

%O 0,4

%A _Paul Barry_, Sep 15 2005