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%I #8 Mar 23 2023 09:25:13
%S 1,2,2,4,4,6,7,8,11,10,16,12,22,14,29,16,37,18,46,20,56,22,67,24,79,
%T 26,92,28,106,30,121,32,137,34,154,36,172,38,191,40,211,42,232,44,254,
%U 46,277,48,301,50,326,52,352,54,379,56,407,58,436,60,466,62,497,64,529,66
%N Expansion of (1+2*x-x^2-2*x^3+x^4) / (1-x^2)^3.
%C Diagonal sums of number triangle A106246. Transform of C(2,n)=(1,2,1,0,0,0,...) under the mapping that takes g(x) to (1/(1-x^2))g(x/(1-x^2)).
%H Colin Barker, <a href="/A106247/b106247.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).
%F a(n)=sum{k=0..floor(n/2), C(n-k, k)C(2, n-2k)}; a(2n)=A000124(n); a(2n+1)=A005843(n+1).
%F From _Colin Barker_, Jul 23 2016: (Start)
%F a(n) = (16+10*n+(-1)^n*(-6+n)*n+n^2)/16.
%F a(n) = (n^2+2*n+8)/8 for n even.
%F a(n) = n+1 for n odd.
%F a(n) = 3*a(n-2)-3*a(n-4)+a(n-6) for n>5.
%F (End)
%t CoefficientList[Series[(1+2x-x^2-2x^3+x^4)/(1-x^2)^3,{x,0,100}],x] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{1,2,2,4,4,6},100] (* _Harvey P. Dale_, Mar 23 2023 *)
%o (PARI) Vec((1+2*x-x^2-2*x^3+x^4)/(1-x^2)^3 + O(x^100)) \\ _Colin Barker_, Jul 23 2016
%K easy,nonn
%O 0,2
%A _Paul Barry_, Apr 26 2005
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