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sum_{k=floor[(n+5)/2] mod 5} C(n,k)
2

%I #9 May 05 2018 09:35:55

%S 0,0,0,0,1,2,7,14,36,72,165,330,715,1430,3004,6008,12393,24786,50559,

%T 101118,204820,409640,826045,1652090,3321891,6643782,13333932,

%U 26667864,53457121,106914242,214146295,428292590,857417220,1714834440,3431847189

%N sum_{k=floor[(n+5)/2] mod 5} C(n,k)

%C Lesser of number of closed walks of length n from a node on a pentagon and number of walks of length n between two adjacent nodes on a pentagon.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2, 3, -6, -1, 2).

%F a(n) =A173125(n)-A000045(n+1). a(2n) =A052964(2n-1); a(2n+1) =A054877(2n+1) =2*a(2n).

%F a(n)= 2*a(n-1) +3*a(n-2) -6*a(n-3) -a(n-4) +2*a(n-5). G.f.: x^4/((1-2*x) * (x^2+x-1) * (x^2-x-1)). [From _R. J. Mathar_, Feb 19 2010]

%e For n=15, k=10 mod 5 gives k=0, 5, 10, or 15, and C(15,0)+C(15,5)+C(15,10)+C(15,15) = 1+3003+3003+1, so a(15)=6008

%t LinearRecurrence[{2,3,-6,-1,2},{0,0,0,0,1},40] (* _Harvey P. Dale_, May 05 2018 *)

%K nonn

%O 0,6

%A _Henry Bottomley_, Feb 10 2010