%I #27 Oct 06 2017 08:36:40

%S 1,1,1,1,1,2,1,2,2,1,3,4,1,3,8,1,4,12,3,1,4,18,8,1,5,24,22,1,5,32,40,

%T 6,1,6,40,73,22,1,6,50,112,66,1,7,60,172,146,10,1,7,72,240,292,48,1,8,

%U 84,335,516,174,1,8,98,440,860,448,20

%N Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X 4 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3), read by rows.

%H Andrew Howroyd, <a href="/A238190/b238190.txt">Table of n, a(n) for n = 3..974</a>

%H Christopher Hunt Gribble, <a href="/A238009/a238009_1.cpp.txt">C++ program</a>

%e The first 13 rows of T(n,k) are:

%e .\ k 0 1 2 3 4 5

%e n

%e 3 1 1

%e 4 1 1

%e 5 1 2

%e 6 1 2 2

%e 7 1 3 4

%e 8 1 3 8

%e 9 1 4 12 3

%e 10 1 4 18 8

%e 11 1 5 24 22

%e 12 1 5 32 40 6

%e 13 1 6 40 73 22

%e 14 1 6 50 112 66

%e 15 1 7 60 172 146 10

%t T[n_, k_] := (2^k Binomial[n - 2k, k] + (Boole[EvenQ[k]] + Boole[OddQ[n] || EvenQ[k]] + Boole[k == 0]) 2^Quotient[k + 1, 2] Binomial[(n - 2k - Mod[n, 2])/2, Quotient[k, 2]])/4; Table[T[n, k], {n, 3, 20}, {k, 0, Floor[n/3]}] // Flatten (* _Jean-François Alcover_, Oct 06 2017, after _Andrew Howroyd_ *)

%o (C++) See Gribble link.

%o (PARI)

%o T(n,k)={(2^k*binomial(n-2*k,k) + ((k%2==0)+(n%2==1||k%2==0)+(k==0)) * 2^((k+1)\2)*binomial((n-2*k-(n%2))/2,k\2))/4}

%o for(n=2,20,for(k=0,floor(n/3), print1(T(n,k), ", "));print) \\ _Andrew Howroyd_, May 29 2017

%Y Cf. A034851, A226048, A102541, A226290, A238009, A228570, A225812, A238189, A228572, A228022, A231145, A231473, A231568, A232440, A228165, A238550, A238551, A238552, A228166, A238555, A238556, A228167, A238557, A238558, A238559, A228168, A238581, A238582, A238583, A228169, A238586, A238592.

%K tabf,nonn

%O 3,6

%A _Christopher Hunt Gribble_, Feb 19 2014

%E Link to C++ program and xrefs updated by _Christopher Hunt Gribble_, Apr 25 2015

%E Terms a(51) and beyond from _Andrew Howroyd_, May 29 2017