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A131816
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Triangle read by rows: A130321 + A059268 - A000012 as infinite lower triangular matrices, where A130321 = (1; 2,1; 4,2,1; ...), A059268 = (1; 1,2; 1,2,4; ...) and A000012 = (1; 1,1; 1,1,1; ...).
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6
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1, 2, 2, 4, 3, 4, 8, 5, 5, 8, 16, 9, 7, 9, 16, 32, 17, 11, 11, 17, 32, 64, 33, 19, 15, 19, 33, 64, 128, 65, 35, 23, 23, 35, 65, 128, 256, 129, 67, 39, 31, 39, 67, 129, 256, 512, 257, 131, 71, 47, 47, 71, 131, 257, 512, 1024, 513, 259, 135, 79, 63, 79, 135, 259, 513, 1024
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OFFSET
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0,2
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COMMENTS
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Row sums = A000295: (1, 4, 11, 26, 57, 120, ...).
If we regard the sequence as an infinite square array read by diagonals then it has the formula U(n,k) = (2^n + 2^k)/2 - 1. This appears to coincide with the number of n X k 0..1 arrays colored with only straight tiles, and new values 0..1 introduced in row major order, i.e., no equal adjacent values form a corner. (Fill the array with 0's and 1's. There must never be 3 adjacent identical values making a corner, only same values in a straight line.) Some solutions with n = k = 4 are:
0 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 1
1 0 1 0 1 0 1 1 1 1 1 0 1 1 0 0 1 0 1 0
1 0 1 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 1
0 1 0 1 1 0 1 1 1 1 1 0 1 1 0 0 0 1 0 1
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LINKS
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FORMULA
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T(n,m) = ((2^(m + 1) - 1) + (2^(n - m + 1) - 1))/2. - Roger L. Bagula, Oct 16 2008
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EXAMPLE
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First few rows of the triangle:
1;
2, 2;
4, 3, 4;
8, 5, 5, 8;
16, 9, 7, 9, 16;
32, 17, 11, 11, 17, 32;
64, 33, 19, 15, 19, 33, 64;
128, 65, 35, 23, 23, 35, 65, 128;
...
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MATHEMATICA
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Table[Table[((2^(m + 1) - 1) + (2^(n - m + 1) - 1))/2, {m, 0, n}], {n, 0, 10}]; Flatten[%] (* Roger L. Bagula, Oct 16 2008 *)
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PROG
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(Haskell)
a131816 n k = a131816_tabl !! n !! k
a131816_row n = a131816_tabl !! n
a131816_tabl = map (map (subtract 1)) $
zipWith (zipWith (+)) a130321_tabl a059268_tabl
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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