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A279212
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Fill an array by antidiagonals upwards; in the top left cell enter a(0)=1; thereafter, in the n-th cell, enter the sum of the entries of those earlier cells that can be seen from that cell.
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10
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1, 1, 2, 2, 6, 11, 4, 15, 39, 72, 8, 37, 119, 293, 543, 16, 88, 330, 976, 2364, 4403, 32, 204, 870, 2944, 8373, 20072, 37527, 64, 464, 2209, 8334, 26683, 74150, 176609, 331072, 128, 1040, 5454, 22579, 79534, 246035, 673156, 1595909, 2997466, 256, 2304, 13176, 59185, 226106, 762221, 2303159, 6231191, 14721429, 27690124
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OFFSET
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0,3
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COMMENTS
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"That can be seen from" means "that are on the same row, column, diagonal, or antidiagonal as".
Conjecture: Every column has a finite number of odd entries, and every row and diagonal have an infinite number of odd entries. - Peter Kagey, Mar 28 2020. The conjecture about columns is true, see that attached pdf file from Alec Jones.
The number of sequences of queen moves from (1, 1) to (n, k) in the first quadrant moving only up, right, diagonally up-right, or diagonally up-left. - Peter Kagey, Apr 12 2020
Column 0 gives A011782. In the column 1, the only powers of 2 occur at positions A233328(k) with value a(k(k+1)/2 + 1), k >=1 (see A335903). Conjecture: Those are the only multiple occurrences of numbers greater than 1 in this sequence (checked through the first 2000 antidiagonals). - Hartmut F. W. Hoft, Jun 29 2020
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LINKS
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FORMULA
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T(0, 0) = 1; T(i, j) = Sum_{k=0..i-1} T(k, j) + Sum_{k=0..j-1} T(i, k) + Sum_{k=0..j-1} T(i+j-k, k) + Sum_{k=0..min(i, j)-1} T(i-k-1, j-k-1), with recursion upwards along antidiagonals. - Hartmut F. W. Hoft, Jun 29 2020
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EXAMPLE
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The array begins:
i/j| 0 1 2 3 4 5 6 7 8
-------------------------------------------------------------
0 | 1 2 11 72 543 4403 37527 331072 2997466 ...
1 | 1 6 39 293 2364 20072 176609 1595909 ...
2 | 2 15 119 976 8373 74150 673156 ...
3 | 4 37 330 2944 26683 246035 ...
4 | 8 88 870 8334 79534 ...
5 | 16 204 2209 22579 ...
6 | 32 464 5454 ...
7 | 64 1040 ...
8 |128 ...
...
For example, when we get to the antidiagonal that reads 4, 15, 39, ..., the reason for the 39 is that from that cell we can see one cell that has been filled in above it (containing 11), one cell to the northwest (2), two cells to the west (1, 6), and two to the southwest (4, 15), for a total of a(8) = 39.
The next pair of duplicates greater than 2 is 2^20 = 1048576 = a(154) = a(231), located in antidiagonals 17 = A233328(2) and 21, respectively. For additional duplicate numbers in this sequence see A335903. - Hartmut F. W. Hoft, Jun 29 2020
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MATHEMATICA
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s[0, 0] = 1; s[i_, j_] := s[i, j] = Sum[s[k, j], {k, 0, i-1}] + Sum[s[i, k], {k, 0, j-1}] + Sum[s[i+j-k, k], {k, 0, j-1}] + Sum[s[i-k-1, j-k-1], {k, 0, Min[i, j] - 1}]
aDiag[m_] := Map[s[m-#, #]&, Range[0, m]]
a279212[n_] := Flatten[Map[aDiag, Range[0, n]]]
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CROSSREFS
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Cf. A064642 is analogous if a cell can only "see" its immediate neighbors.
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KEYWORD
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AUTHOR
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STATUS
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approved
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