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A175899 a(n) = a(n-2) + a(n-3) + 2*a(n-4), with a(1) = 0, a(2) = 2, a(3) = 3, a(4) = 10.

%I A175899 #28 Feb 01 2018 03:43:06
%S A175899 0,2,3,10,5,17,21,42,48,97,132,229,325,555,818,1338,2023,3266,4997,
%T A175899 7965,12309,19494,30268,47733,74380,116989,182649,286835,448398,
%U A175899 703462,1100531,1725530,2700789,4232985,6627381,10384834,16261944,25478185,39901540,62509797
%N A175899 a(n) = a(n-2) + a(n-3) + 2*a(n-4), with a(1) = 0, a(2) = 2, a(3) = 3, a(4) = 10.
%C A175899 According to the reference, p divides a(p) for every prime p.
%H A175899 Reinhard Zumkeller, <a href="/A175899/b175899.txt">Table of n, a(n) for n = 1..1000</a>
%H A175899 Eric Pite, <a href="http://www.jstor.org/stable/10.4169/002557010x521877">Problem 1851</a>, Mathematics Magazine 83 (2010) 303.
%H A175899 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 1, 2).
%F A175899 G.f.: x*(-2*x-3*x^2-8*x^3)/(-1+x^2+x^3+2*x^4). - _Harvey P. Dale_, Jul 24 2011
%F A175899 a(n) = n*sum(k=1..n/2, sum(j=0..k, binomial(j,n-4*k+2*j)*2^(k-j) * binomial(k,j))/k), n>0. - _Vladimir Kruchinin_, Oct 21 2011
%p A175899 a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <2|1|1|0>>^n.
%p A175899         <<4,0,2,3>>)[1, 1]:
%p A175899 seq(a(n), n=1..50);  # _Alois P. Heinz_, Oct 21 2011
%t A175899 LinearRecurrence[{0,1,1,2},{0,2,3,10},40] (* _Harvey P. Dale_, Jul 24 2011 *)
%o A175899 (Maxima) a(n):=n*sum(sum(binomial(j,n-4*k+2*j)*2^(k-j)*binomial(k,j), j,0,k)/k, k,1,n/2); /* _Vladimir Kruchinin_, Oct 21 2011 */
%o A175899 (Haskell)
%o A175899 a175899 n = a175899_list !! (n-1)
%o A175899 a175899_list = 0 : 2 : 3 : 10 : zipWith (+) (map (* 2) a175899_list)
%o A175899    (zipWith (+) (tail a175899_list) (drop 2 a175899_list))
%o A175899 -- _Reinhard Zumkeller_, Mar 23 2012
%Y A175899 Cf. A001608, A001634.
%K A175899 nonn,easy
%O A175899 1,2
%A A175899 _John W. Layman_, Oct 11 2010