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A244051 Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order. 13
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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
LINKS
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
Sequence in context: A167269 A105535 A182980 * A207974 A239454 A108888
KEYWORD
nonn,tabf,easy
AUTHOR
Omar E. Pol, Nov 08 2014
STATUS
approved

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Last modified July 23 23:47 EDT 2024. Contains 374575 sequences. (Running on oeis4.)