%I #15 May 03 2022 18:52:57

%S 0,5,3,25,25,131,175,705,1137,3875,7095,21649,43225,122435,259423,

%T 698625,1541985,4011971,9107175,23143825,53559817,133933475,314086735,

%U 776787009,1838300625,4512108515,10745077143,26237143825,62749602745

%N a(n) = A001333(n) - (-2)^(n-1), n > 0.

%C Conjecture: for odd primes p, p divides a(p). Note that (a(n)) and A001333 have different offsets.

%C The conjecture follows from the formula A001333(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2. - _Max Alekseyev_, Oct 16 2005

%H Colin Barker, <a href="/A111108/b111108.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,5,2).

%F From _Colin Barker_, Apr 30 2019: (Start)

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

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

%F (End)

%t LinearRecurrence[{0,5,2},{0,5,3},30] (* _Harvey P. Dale_, May 03 2022 *)

%o (PARI) concat(0, Vec(x^2*(5 + 3*x) / ((1 + 2*x)*(1 - 2*x - x^2)) + O(x^35))) \\ _Colin Barker_, May 01 2019

%Y Cf. A001333, A000204.

%K easy,nonn

%O 1,2

%A _Creighton Dement_, Oct 14 2005