A302247 Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nondecreasing order.
1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Examples
Triangle begins: 1; 1,1,2; 1,1,1,1,2,3; 1,1,1,1,1,1,1,2,2,2,3,4; 1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5; 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,4,5,6; ... For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There are seven 1's, three 2's, only one 3 and only one 4, so the 4th row of this triangle is [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4]. 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 nondecreasing 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
Mirror of A302246.
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).
Programs
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Mathematica
nrows=10; Array[Sort[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)); \\ Michel Marcus, Jun 16 2022
Comments