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A231568
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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.
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21
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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, 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, 98, 335, 292, 10, 1, 9, 112, 440, 516, 48, 1, 10, 128, 578, 860, 174
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OFFSET
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4,6
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LINKS
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EXAMPLE
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The first 14 rows of T(n,k) are:
.\ k 0 1 2 3 4
n
4 1 1
5 1 1
6 1 2
7 1 2
8 1 3 2
9 1 3 4
10 1 4 8
11 1 4 12
12 1 5 18 3
13 1 5 24 8
14 1 6 32 22
15 1 6 40 40
16 1 7 50 73 6
17 1 7 60 112 22
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MATHEMATICA
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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 *)
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PROG
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(C++) See Gribble link.
(PARI)
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}
for(n=2, 20, for(k=0, floor(n/4), print1(T(n, k), ", ")); print) \\ Andrew Howroyd, May 29 2017
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CROSSREFS
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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.
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KEYWORD
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tabf,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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