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A173997
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Irregular triangle by columns derived from (1, 2, 3, ...) * (1, 2, 3, ...).
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4
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1, 2, 3, 2, 4, 4, 5, 6, 3, 6, 8, 6, 7, 10, 9, 4, 8, 12, 12, 8, 9, 14, 15, 12, 5, 10, 16, 18, 16, 10, 11, 18, 21, 20, 15, 6, 12, 20, 24, 24, 20, 12, 13, 22, 27, 28, 25, 18, 7, 14, 24, 30, 32, 30, 24, 14, 15, 26, 33, 36, 35, 30, 21, 8, 16, 28, 36, 40, 40, 36, 28, 16
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OFFSET
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1,2
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COMMENTS
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Given a (1, 2, 3, ...) * (1, 2, 3, ...) multiplication table; leftmost column of the triangle = (1, 2, 3, ...). Then shift down each successive column of the array twice to get this irregular triangle.
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LINKS
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FORMULA
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T(n, k) = k*(2 - 2*k + n), with 1 <= k <= floor((n + 1)/2). - Stefano Spezia, Apr 19 2022
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EXAMPLE
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Given:
1, 2, 3, 4, 5, ...
2, 4, 6, 8, 10, ...
3, 6, 9, 12, 15, ...
4, 8, 12, 16, 20, ...
...
After the shift twice operation, we obtain:
1;
2;
3, 2;
4, 4;
5, 6, 3;
6, 8, 6;
7, 10, 9, 4;
8, 12, 12, 8;
9, 14, 15, 12, 5;
10, 16, 18, 16, 10;
11, 18, 21, 20, 15, 6;
12, 20, 24, 24, 20, 12;
...
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MATHEMATICA
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Flatten[Table[k(2-2k+n), {n, 16}, {k, Floor[(n+1)/2]}]] (* Stefano Spezia, Apr 19 2022 *)
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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STATUS
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approved
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