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a(n) = 20*binomial(n,6) + 2*binomial(n,3) + 1.
2

%I #9 Mar 07 2021 19:49:10

%S 1,1,1,3,9,21,61,211,673,1849,4441,9571,18921,34893,60789,101011,

%T 161281,248881,372913,544579,777481,1087941,1495341,2022483,2695969,

%U 3546601,4609801,5926051,7541353,9507709,11883621,14734611,18133761,22162273,26910049,32476291

%N a(n) = 20*binomial(n,6) + 2*binomial(n,3) + 1.

%C a(n) is the number of ternary strings of length n that contain either none or three 0's and either none or three 1's.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F E.g.f.: exp(x)*(1 + x^3/6)^2.

%F O.g.f.:(1 - 6*x + 15*x^2 - 18*x^3 + 9*x^4 + 19*x^6)/(1 - x)^7. - _Stefano Spezia_, Feb 19 2021

%e a(7)=211 since the strings are the 140 permutations of 0001112, the 35 permutations of 0002222, the 35 permutations of 1112222, and 2222222.

%Y Cf. A341703, A341705.

%K nonn,easy

%O 0,4

%A _Enrique Navarrete_, Feb 17 2021