%I #16 Dec 12 2024 14:55:31
%S 4,30,106,292,712,1618,3518,7432,15404,31526,63986,129164,259824,
%T 521498,1045254,2093232,4189716,8383278,16771066,33547380,67100824,
%U 134208610,268425166,536859352,1073728892,2147469238,4294951298,8589916892
%N Number of n X 3 1..2 arrays containing at least one of each value, and all equal values connected.
%H R. H. Hardin, <a href="/A166761/b166761.txt">Table of n, a(n) for n=1..41</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,16,-9,2)
%F Empirical: a(n) = 6*a(n-1) - 14*a(n-2) + 16*a(n-3) -9*a(n-4) + 2*a(n-5).
%F From _G. C. Greubel_, May 26 2016: (Start)
%F Empirical a(n) = (3*2^(n + 5) - 2*n^3 - 9*n^2 - 73*n - 96)/3.
%F Empirical G.f.: (1/3)*( 96/(1 - 2*x) + 6*(-16 + 34*x - 25*x^2 + 5*x^3)/(1 - x)^4 ).
%F Empirical E.g.f.: (1/3)*(96*exp(x) - (96 + 84*x + 15*x^2 + 2*x^3 ) )*exp(x). (End)
%F From _Andrew Howroyd_, Dec 12 2024: (Start)
%F The above empirical formulas are correct.
%F a(n) = 2*A378933(n).
%F (End)
%e Some solutions for n=4
%e ...2.2.2...2.2.2...1.1.2...1.1.1...2.1.1...1.1.1...1.2.2...1.2.2...1.1.1
%e ...2.1.1...1.2.2...1.2.2...2.2.1...2.2.2...2.1.1...1.2.2...1.2.2...1.2.1
%e ...2.1.1...1.2.1...1.2.2...2.2.1...2.2.2...2.2.1...1.1.2...1.1.2...1.2.1
%e ...2.2.1...1.1.1...1.1.2...2.2.1...2.2.2...2.2.2...2.2.2...1.1.1...1.1.1
%e ------
%e ...1.2.2...2.2.2...1.1.2...1.1.1...1.1.1...1.1.2...1.2.2...1.2.2...1.1.1
%e ...1.1.2...2.1.2...1.1.2...2.2.1...1.1.2...1.1.2...1.1.2...1.2.1...1.1.1
%e ...1.1.2...1.1.1...1.2.2...2.2.1...1.1.2...1.1.2...1.1.2...1.1.1...2.2.1
%e ...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...1.1.1...2.2.2...1.1.1...2.1.1
%Y Twice row 3 of A378932.
%Y Cf. A378933.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 21 2009