OFFSET
1,2
COMMENTS
Row n has length sigma(n) = A000203(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
LINKS
Samuel Harkness, Table of n, a(n) for n = 1..10000
Samuel Harkness, Log-log Scatterplot of the first 1000000 terms
EXAMPLE
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)
MATHEMATICA
A244051row[n_]:=Flatten[Map[ConstantArray[#, n/#]&, Reverse[Divisors[n]]]];
Array[A244051row, 10] (* Paolo Xausa, Oct 16 2023 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Omar E. Pol, Nov 08 2014
STATUS
approved