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A309038
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Irregular triangle T read by rows: given a square made of n^2 squares of unit area, T(n, k) is the longest perimeter that can be obtained by removing k of n^2 squares such that the modified figure remains connected and without holes (n >= 0 and 0 <= k <= n^2).
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4
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0, 4, 0, 8, 8, 8, 4, 0, 12, 14, 16, 18, 20, 16, 12, 8, 4, 0, 16, 18, 20, 22, 24, 26, 28, 28, 28, 26, 24, 20, 16, 12, 8, 4, 0, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 42, 40, 38, 36, 32, 28, 24, 20, 16, 12, 8, 4, 0, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 56, 56, 56, 56, 56, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20, 16, 12, 8, 4, 0
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OFFSET
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0,2
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COMMENTS
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All the terms of this sequence are even numbers (A005843).
In the figure, two unit area squares can be connected in a corner or sideways.
Every n-th row of the triangle is made of almost four successive finite arithmetic progressions characterized respectively by the following common differences: 2, 0, -2, -4. If we let h_i(n) be the number of first differences of i-th progression (i = 1,2,3,4), we have that 4*n + 2*h_1(n) - 2*h_3(n) - 4*h_4(n) = 0 and h_1(n) + h_2(n) + h_3(n) + h_4(n) = n^2.
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LINKS
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FORMULA
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T(n, k) = 2*(2*n + k) for 0 <= k <= h_1(n), T(n, k) = 2*(2*n + h_1(n)) for h_1(n) <= k <= h_1(n) + h_2(n), T(n, k) = 2*(2*(n + h_1(n)) + h_2(n) - k) for h_1(n) + h_2(n) <= k <= h_1(n) + h_2(n) + h_3(n), T(n, k) = 2*(2*(n + h_2(n) - k) + 3*h_1(n) + h_3(n)), for h_1(n) + h_2(n) + h_3(n) <= k <= n^2, where h_4(n) = n for 0 <= n <= 2 and h_4(n) = (1/8)*(-29 + 12*n + 2*n^2 - 3*cos(n*Pi) - 12*sin(n*Pi/2)) for n > 2, h_3(n) = 2*delta(n, 4) - 4*delta(n, 1) + 1 - cos(n*Pi) + 2*sin(n*Pi/2) and delta(i, j) is the Kronecker delta, h_2(n) = 2*(delta(n, 2) + delta(n, 4)) for 0 <= n <= 4 and h_2(n) = (1/8)*(71 - 20*n + 2*n^2 + 25*cos(n*Pi) + 4*sin(n*Pi/2)) for n > 4, h_1(n) = n^2 - (h_1(n) + h_2(n) + h_3(n)).
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EXAMPLE
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The triangle T(n, k) begins:
---+-------------------------------------------------------------------
n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
---+-------------------------------------------------------------------
0 | 0
1 | 4 0
2 | 8 8 8 4 0
3 | 12 14 16 18 20 16 12 8 4 0
4 | 16 18 20 22 24 26 28 28 28 26 24 20 16 12 8 4 0
...
Here are the values of h_i's for the first seven rows of the triangle T:
n h_1(n) h_2(n) h_3(n) h_4(n)
--------------------------------------
0 0 0 0 0
1 0 0 0 1
2 0 2 0 2
3 4 0 0 5
4 6 2 2 6
5 12 0 4 9
6 16 6 0 14
...
Illustrations for n = 4, k=0..15 by Andrew Howroyd, Sep 01 2019: (Start)
__.__.__.__ __.__.__.__ __.__.__.__ __.__.__.__
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(16) (18) (20) (22)
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(24) (26) (28) (28)
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(28) (26) (24) (20)
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(16) (12) (8) (4)
(End)
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MATHEMATICA
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h4[n_]:=If[n>2, (1/8)(-29+12n+2n^2-3*Cos[n*Pi]-12*Sin[n*Pi/2]), n]; h3[n_]:=1-Cos[n*Pi]-4*KroneckerDelta[n, 1]+2*KroneckerDelta[n, 4]+2*Sin[n*Pi/2]; h2[n_]:=If[n>4, (1/8)(71-20n+2n^2+25Cos[n*Pi]+4Sin[n*Pi/2]), 2*(KroneckerDelta[n, 2]+KroneckerDelta[n, 4])]; h1[n_]:=n^2-(h2[n]+h3[n]+h4[n]); T[n_, k_]:=If[0<=k<=h1[n], 2(2n+k), If[h1[n]<k<=(h1[n]+h2[n]), 2(2n+h1[n]), If[(h1[n]+h2[n])<k<=(h1[n]+h2[n]+h3[n]), 2(2(n+h1[n])+h2[n]-k), 2(2(n+h2[n]-k)+3h1[n]+h3[n])]]]; Flatten[Table[T[n, k], {n, 0, 6}, {k, 0, n^2}]]
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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