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%I #11 Mar 07 2021 19:49:17
%S 1,1,3,7,19,51,121,253,477,829,1351,2091,3103,4447,6189,8401,11161,
%T 14553,18667,23599,29451,36331,44353,53637,64309,76501,90351,106003,
%U 123607,143319,165301,189721,216753,246577,279379,315351,354691,397603,444297,494989,549901
%N a(n) = 6*binomial(n,4) + 2*binomial(n,2) + 1.
%C a(n) is the number of ternary strings of length n that contain either none or two 0's and either none or two 1's.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F E.g.f.: exp(x)*(1 + x^2/2)^2.
%F From _Stefano Spezia_, Feb 19 2021: (Start)
%F O.g.f.:(1 - 4*x + 8*x^2 - 8*x^3 + 9*x^4)/(1 - x)^5.
%F a(n) = (4 - 10*n + 15*n^2 - 6*n^3 + n^4)/4. (End)
%F a(n) = 2*A004255(n-1) + 1. - _Hugo Pfoertner_, Feb 19 2021
%e a(6)=121 since the strings are the 90 permutations of 110022, the 15 permutations of 002222, the 15 permutations of 112222, and 222222.
%Y Cf. A004255, A341704, A341705.
%K nonn,easy
%O 0,3
%A _Enrique Navarrete_, Feb 17 2021
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