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%I #37 Oct 06 2017 08:36:26
%S 1,1,1,1,1,2,1,2,1,3,2,1,3,4,1,4,8,1,4,12,1,5,18,3,1,5,24,8,1,6,32,22,
%T 1,6,40,40,1,7,50,73,6,1,7,60,112,22,1,8,72,172,66,1,8,84,240,146,1,9,
%U 98,335,292,10,1,9,112,440,516,48,1,10,128,578,860,174
%N Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X 5 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4), read by rows.
%H Andrew Howroyd, <a href="/A231568/b231568.txt">Table of n, a(n) for n = 4..989</a>
%H Christopher Hunt Gribble, <a href="/A238009/a238009_1.cpp.txt">C++ program</a>
%e The first 14 rows of T(n,k) are:
%e .\ k 0 1 2 3 4
%e n
%e 4 1 1
%e 5 1 1
%e 6 1 2
%e 7 1 2
%e 8 1 3 2
%e 9 1 3 4
%e 10 1 4 8
%e 11 1 4 12
%e 12 1 5 18 3
%e 13 1 5 24 8
%e 14 1 6 32 22
%e 15 1 6 40 40
%e 16 1 7 50 73 6
%e 17 1 7 60 112 22
%t T[n_, k_] := (2^k Binomial[n - 3k, k] + (Boole[EvenQ[k]] + Boole[EvenQ[n] || EvenQ[k]] + Boole[k == 0]) 2^Quotient[k+1, 2] Binomial[(n - 3k - Mod[k, 2] - Mod[n, 2])/2, Quotient[k, 2]])/4; Table[T[n, k], {n, 4, 20}, {k, 0, Floor[n/4]}] // 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-3*k,k) + ((k%2==0)+(n%2==0||k%2==0)+(k==0)) * 2^((k+1)\2)*binomial((n-3*k-(k%2)-(n%2))/2,k\2))/4}
%o for(n=2,20,for(k=0,floor(n/4), print1(T(n,k), ", "));print) \\ _Andrew Howroyd_, May 29 2017
%Y Cf. A034851, A226048, A102541, A226290, A238009, A228570, A225812, A238189, A238190, A228572, A228022, A231145, A231473, A232440, A228165, A238550, A238551, A238552, A228166, A238555, A238556, A228167, A238557, A238558, A238559, A228168, A238581, A238582, A238583, A228169, A238586, A238592.
%K tabf,nonn
%O 4,6
%A _Christopher Hunt Gribble_, Feb 23 2014
%E Terms extended and xrefs updated by _Christopher Hunt Gribble_, Apr 26 2015
%E Terms a(32) and beyond from _Andrew Howroyd_, May 29 2017
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