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A359279
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Irregular triangle T(n,k) (n>=1, k>=1) read by rows in which the length of row n equals the partition number A000041(n-1) and every column k gives the positive triangular numbers A000217.
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2
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1, 3, 6, 1, 10, 3, 1, 15, 6, 3, 1, 1, 21, 10, 6, 3, 3, 1, 1, 28, 15, 10, 6, 6, 3, 3, 1, 1, 1, 1, 36, 21, 15, 10, 10, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 55, 36, 28, 21, 21, 15, 15, 10, 10, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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1,2
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COMMENTS
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All divisors of the largest partition into consecutive parts of all terms in row n are also all parts of all partitions of n.
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins:
1;
3;
6, 1;
10, 3, 1;
15, 6, 3, 1, 1;
21, 10, 6, 3, 3, 1, 1;
28, 15, 10, 6, 6, 3, 3, 1, 1, 1, 1;
36, 21, 15, 10, 10, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1;
45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
For n = 4 the fourth row is [10, 3, 1]. The largest partition into consecutive parts of every term are respectively [4, 3, 2, 1], [2, 1], [1]. The divisors of these parts are [(1, 2, 4), (1, 3), (1, 2), (1)], [(1, 2), (1)], [1]. These 12 divisors are also all parts of all partitions of 4. They are [(4), (2, 2), (3, 1), (2, 1, 1), (1, 1, 1, 1)]. (End)
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MATHEMATICA
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A359279[rowmax_]:=Table[Flatten[Table[ConstantArray[(n-m)(n-m+1)/2, PartitionsP[m]-PartitionsP[m-1]], {m, 0, n-1}]], {n, rowmax}];
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PROG
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(PARI)
A359279(rowmax)=vector(rowmax, n, concat(vector(n, m, vector(numbpart(m-1)-numbpart(m-2), i, (n-m+1)*(n-m+2)/2))));
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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