A244051 Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order.

1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 6, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 8, 4, 4, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 9, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 5, 5, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Omar E. Pol, Nov 08 2014

Keywords

• nonn
• tabf
• easy

Comments

Row n has length sigma(n) = A000203(n).

Row sums give n*A000005(n) = A038040(n).

Column 1 is A000027.

Both columns 2 and 3 are A032742, n > 1.

For any k > 0 and t > 0, the sequence contains exactly one run of k consecutive t's. - Rémy Sigrist, Feb 11 2019

From Omar E. Pol, Dec 04 2019: (Start)

The number of parts congruent to 0 (mod m) in row m*n equals sigma(n) = A000203(n).

The number of parts greater than 1 in row n equals A001065(n), the sum of aliquot parts of n.

The number of parts greater than 1 and less than n in row n equals A048050(n), the sum of divisors of n except for 1 and n.

The number of partitions in row n equals A000005(n), the number of divisors of n.

The number of partitions in row n with an odd number of parts equals A001227(n).

The sum of odd parts in row n equals the sum of parts of the partitions in row n that have an odd number of parts, and equals the sum of all parts in the partitions of n into consecutive parts, and equals A245579(n) = n*A001227(n).

The decreasing records in row n give the n-th row of A056538.

Row n has n 1's which are all at the end of the row.

First n rows contain A000217(n) 1's.

The number of k's in row n is A126988(n,k).

The number of odd parts in row n is A002131(n).

The k-th block in row n has A027750(n,k) parts.

Right border gives A000012. (End)

The r-th row of the triangle begins at index k = A160664(r-1). - Samuel Harkness, Jun 21 2022

Examples

			Triangle begins:
   [1];
   [2], [1,1];
   [3], [1,1,1];
   [4], [2,2], [1,1,1,1];
   [5], [1,1,1,1,1];
   [6], [3,3], [2,2,2], [1,1,1,1,1,1];
   [7], [1,1,1,1,1,1,1];
   [8], [4,4], [2,2,2,2], [1,1,1,1,1,1,1,1];
   [9], [3,3,3], [1,1,1,1,1,1,1,1,1];
  [10], [5,5], [2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1];
  [11], [1,1,1,1,1,1,1,1,1,1,1];
  [12], [6,6], [4,4,4], [3,3,3,3], [2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1];
  [13], [1,1,1,1,1,1,1,1,1,1,1,1,1];
  [14], [7,7], [2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1];
  [15], [5,5,5], [3,3,3,3,3], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];
  [16], [8,8], [4,4,4,4], [2,2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];
  ...
For n = 6 the 11 partitions of 6 are [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. There are only four partitions of 6 that contain equal parts so the 6th row of triangle is [6], [3, 3], [2, 2, 2], [1, 1, 1, 1, 1, 1]. The number of parts equals sigma(6) = A000203(6) = 12. The row sum is A038040(6) = 6*A000005(6) = 6*4 = 24.
From _Omar E. Pol_, Dec 04 2019: (Start)
The structure of the above triangle is as follows:
   1;
   2 11;
   3    111;
   4 22     1111;
   5             11111;
   6 33 222            111111;
   7                          1111111;
   8 44     2222                      11111111;
   9    333                                    111111111;
  ... (End)
		

Links

• Samuel Harkness, Table of n, a(n) for n = 1..10000
• Samuel Harkness, Log-log Scatterplot of the first 1000000 terms

Crossrefs

Cf. A000005, A000012, A000041, A000203, A000217, A001065, A001227, A002131, A027750, A032742, A038040, A048050, A056538, A126988, A160664, A237593, A245579, A299765, A328365, A309400 (mirror).

Programs

Mathematica

A244051row[n_]:=Flatten[Map[ConstantArray[#,n/#]&,Reverse[Divisors[n]]]];
Array[A244051row,10] (* Paolo Xausa, Oct 16 2023 *)

PARI

tabf(nn) = {for (n=1, nn, d = Vecrev(divisors(n)); for (i=1, #d, for (j=1, n/d[i], print1(d[i], ", "));); print(););} \\ Michel Marcus, Nov 08 2014