A336811 Irregular triangle read by rows T(n,k) in which the length of row n equals the partition number A000041(n-1) and every column k gives the positive integers A000027, with n >= 1 and k >= 1.

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

Offset

1,2

Comments

In other words: row n lists A028310(n-1) blocks where the m-th block consists of A187219(m) copies of n - m + [m=1], with n >= 1 and m >= 1, where [] is the Iverson bracket. [Corrected by Paolo Xausa, Feb 10 2023]

All divisors of all terms in row n are also all parts in the last section of the set of partitions of n.

Thus all divisors of all terms of the first n rows of triangle are also all parts of all partitions of n. In other words: all divisors of the first A000070(n-1) terms of the sequence are also all parts of all partitions of n. - Omar E. Pol, Jun 19 2021

From Omar E. Pol, Jul 31 2021: (Start)

The number of k's in row n is equal to A002865(n-k), 1 <= k <= n.

The number of terms >= k in row n is equal to A000041(n-k), 1 <= k <= n.

The number of k's in the first n rows (or in the first A000070(n-1) terms of the sequence) is equal to A000041(n-k), 1 <= k <= n.

The number of terms >= k in the first n rows (or in the first A000070(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.

First n rows of triangle (or first A000070(n-1) terms of the sequence) in nonincreasing order give the n-th row of A176206. (End)

Links

Paolo Xausa, Table of n, a(n) for n = 1..11732 (rows 1..27 of triangle, flattened).

Example

Triangle begins:
1;
2;
3, 1;
4, 2, 1;
5, 3, 2, 1, 1;
6, 4, 3, 2, 2, 1, 1;
7, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1;
8, 6, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1;
9, 7, 6, 5, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
...
For n = 6, by definition the length of row 6 is A000041(6-1) = A000041(5) = 7, so the row 6 of triangle has seven terms. Since every column lists the positive integers A000027 so the row 6 is [6, 4, 3, 2, 2, 1, 1].
Then we have that the divisors of the numbers of the 6th row are:
.
6th row of the triangle ---------->   6 4 3 2 2 1 1
                                      3 2 1 1 1
                                      2 1
                                      1
.
There are seven 1's, four 2's, two 3's, one 4 and one 6.
In total there are 7 + 4 + 2 + 1 + 1 = 15 divisors.
On the other hand the last section of the set of the partitions of 6 can be represented in several ways, five of them as shown below:
._ _ _ _ _ _
|_ _ _      |       6    6                  6                       6
|_ _ _|_    |     3 3    3 3              3   3                     3   3
|_ _    |   |     4 2    4 2            4       2                     4     2
|_ _|_ _|_  |   2 2 2    2 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           1
          | |       1            1                            1       1
          |_|       1              1                            1   1
.
   Figure 1.  Figure 2.  Figure 3.        Figure 4.                   Figure 5.
.
In every figure there are seven 1's, four 2's, two 3's, one 4 and one 6, as shown also the 6th row of A182703.
In total there are 7 + 4 + 2 + 1 + 1 = A138137(6) = 15 parts in every figure.
Figure 5 is an arrangement that shows the correspondence between divisors and parts since the columns give the divisors of the terms of 6th row of triangle.
Finally we can see that all divisors of all numbers in the 6th row of the triangle are the same positive integers as all parts in the last section of the set of the partitions of 6.
Example edited by Omar E. Pol, Aug 10 2021

Mathematica

A336811[row_]:=Flatten[Table[ConstantArray[row-m, PartitionsP[m]-PartitionsP[m-1]], {m, 0, row-1}]];
Array[A336811, 10] (* Generates 10 rows *) (* Paolo Xausa, Feb 10 2023 *)

Prog

(PARI) f(n) = numbpart(n-1);
T(n, k) = {if (k > f(n), error("invalid k")); if (k==1, return (n)); my(s=0); while (k <= f(n-1), s++; n--; ); 1+s; }
tabf(nn) = {for (n=1, nn, for (k=1, f(n), print1(T(n, k), ", "); ); print; ); } \\ Michel Marcus, Jan 13 2021

Crossrefs

Row sums give A000070.

Row n has length A000041(n-1).

Every column k gives A000027.

Companion of A176206.

Cf. A000007, A000041, A027750, A028310, A002865, A133735, A135010, A138121, A138137, A182703, A187219, A207378, A221529, A336812, A339278, A340035, A340061, A346741.

Sequence in context: A195164 A223025 A087088 * A255250 A273620 A104705

Adjacent sequences: A336808 A336809 A336810 * A336812 A336813 A336814

Keyword

nonn,tabf

Author

Omar E. Pol, Nov 20 2020

Status

approved