A302246 Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nonincreasing order.
1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1
Offset: 1
Examples
Triangle begins: 1; 2,1,1; 3,2,1,1,1,1; 4,3,2,2,2,1,1,1,1,1,1,1; 5,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1; 6,5,4,4,3,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1; ... For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There is only one 4, only one 3, three 2's and seven 1's, so the 4th row of this triangle is [4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1]. On the other hand for n = 4 the 4th row of A176206 is [4, 3, 2, 2, 1, 1, 1] and the divisors of these terms are [1, 2, 4], [1, 3], [1, 2], [1, 2], [1], [1], [1] the same as the 4th row of A338156. These divisors listed in nonincreasing order give the 4th row of this triangle. - _Omar E. Pol_, Jun 16 2022
Links
- Paolo Xausa, Table of n, a(n) for n = 1..9687, (rows 1..18 of triangle, flattened).
Crossrefs
Both column 1 and 2 are A000027.
Row n has length A006128(n).
The sum of row n is A066186(n).
The number of parts k in row n is A066633(n,k).
The sum of all parts k in row n is A138785(n,k).
The number of parts >= k in row n is A181187(n,k).
The sum of all parts >= k in row n is A206561(n,k).
The number of parts <= k in row n is A210947(n,k).
The sum of all parts <= k in row n is A210948(n,k).
Cf. A302247 (mirror).
Programs
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Mathematica
nrows=10;Array[ReverseSort[Flatten[IntegerPartitions[#]]]&,nrows] (* Paolo Xausa, Jun 16 2022 *)
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PARI
row(n) = my(list = List()); forpart(p=n, for (k=1, #p, listput(list, p[k]));); vecsort(Vec(list), , 4); \\ Michel Marcus, Jun 16 2022
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