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A209490
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Table T(d,n), read by antidiagonals, gives the number of subsets of length n containing an arithmetic progression of length 3 with distance d.
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3
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1, 0, 3, 0, 0, 8, 0, 0, 4, 20, 0, 0, 0, 15, 47, 0, 0, 0, 0, 37, 107, 0, 0, 0, 0, 16, 87, 238, 0, 0, 0, 0, 0, 60, 200, 520, 0, 0, 0, 0, 0, 0, 169, 448, 1121, 0, 0, 0, 0, 0, 0, 64, 387, 992, 2391, 0, 0, 0, 0, 0, 0, 0, 240, 865, 2160, 5056
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OFFSET
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3,3
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COMMENTS
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Offset for n is 3, offset for d is 1, thus 1st entry is T(1,3).
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LINKS
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FORMULA
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T(d,n) = 2^n - prod_{1=0 to d-1} Tri(floor((n + i)/d) + 2) where Tri(n) is the n-th tribonacci number.
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EXAMPLE
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Table begins:
1, 3, 8, 20, 47, 107, 238, 520 ...
0, 0, 4, 15, 37, 87, 200, 448 ...
0, 0, 0, 0, 16, 60, 169, 387 ...
0, 0, 0, 0, 0, 0, 64, 240 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
0, 0, 0, 0, 0, 0, 0, 0 ...
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For T(2,5) we count subsets of {1,...,5} containing {1,3,5}, the only d=2 AP possible here. There are 4 subsets containing {1,3,5} so T(2,5) = 4.
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MATHEMATICA
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T[0]=0; T[1] = 1; T[2] = 1; T[n_] := T[n - 1] + T[n - 2] + T[n - 3]; a[d_, n_] := 2^n - Product[T[Floor[(n + i)/d] + 2], {i, 0, d - 1}]; Table[a[i, j], {i, 1, 10}, {j, 3, 10}]; Flatten[Table[a[j - i + 1, i + 3], {j, 0, 10}, {i, 0, j}]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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