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A135765
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Distribute the odd numbers in columns based on the occurrence of "3" in each prime factorization; square array A(row, col) = 3^(row-1) * A007310(col), read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...
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10
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1, 5, 3, 7, 15, 9, 11, 21, 45, 27, 13, 33, 63, 135, 81, 17, 39, 99, 189, 405, 243, 19, 51, 117, 297, 567, 1215, 729, 23, 57, 153, 351, 891, 1701, 3645, 2187, 25, 69, 171, 459, 1053, 2673, 5103, 10935, 6561, 29, 75, 207, 513, 1377, 3159, 8019, 15309, 32805
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OFFSET
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1,2
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COMMENTS
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The Table can be constructed by multiplying sequence A000244 by A007310.
A permutation of odd numbers. Adding one to each term and then dividing by two gives a related table A254051, which for any odd number, located in this array as x = A(row,col), gives the result at A254051(row+1,col) after one combined Collatz step (3x+1)/2 -> x (A165355) has been applied.
(End)
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LINKS
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FORMULA
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With both row and col starting from 1:
Above in array form:
(End)
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EXAMPLE
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The top left corner of the array:
1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, ...
3, 15, 21, 33, 39, 51, 57, 69, 75, 87, 93, 105, ...
9, 45, 63, 99, 117, 153, 171, 207, 225, 261, 279, 315, ...
27, 135, 189, 297, 351, 459, 513, 621, 675, 783, 837, 945, ...
81, 405, 567, 891, 1053, 1377, 1539, 1863, 2025, 2349, 2511, 2835, ...
243, 1215, 1701, 2673, 3159, 4131, 4617, 5589, 6075, 7047, 7533, 8505, ...
etc.
For n = 6, we have [A002260(6), A004736(6)] = [3, 1] (that is 6 corresponds to location 3,1 (row,col) in above table) and A(3,1) = A000244(3-1) * A007310(1) = 3^2 * 1 = 9.
For n = 9, we have [A002260(9), A004736(9)] = [3, 2] (9 corresponds to location 3,2) and A(3,2) = A000244(3-1) * A007310(2) = 3^2 * 5 = 9*5 = 45.
For n = 13, we have [A002260(13), A004736(13)] = [3, 3] (13 corresponds to location 3,3) and A(3,3) = A000244(3-1) * A007310(3) = 3^2 * 7 = 9*7 = 63.
For n = 23, we have [A002260(23), A004736(23)] = [2, 6] (23 corresponds to location 2,6) and A(2,6) = A000244(2-1) * A007310(6) = 3^1 * 17 = 51.
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MAPLE
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N:= 20:
B:= [seq(op([6*n+1, 6*n+5]), n=0..floor((N-1)/2))]:
[seq(seq(3^j*B[i-j], j=0..i-1), i=1..N)]; # Robert Israel, Jan 26 2015
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PROG
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(Scheme, two versions)
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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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