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A340581 Irregular triangle read by rows in which row n has length A014153(n-1) and every column k lists the positive integers A000027, n >= 1, k >= 1.

%I A340581 #43 Sep 04 2023 12:10:40
%S A340581 1,2,1,1,3,2,2,1,1,1,1,4,3,3,2,2,2,2,1,1,1,1,1,1,1,5,4,4,3,3,3,3,2,2,
%T A340581 2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,6,5,5,4,4,4,4,3,3,3,3,3,3,3,2,2,2,
%U A340581 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
%N A340581 Irregular triangle read by rows in which row n has length A014153(n-1) and every column k lists the positive integers A000027, n >= 1, k >= 1.
%C A340581 Row n lists in nonincreasing order the first A014153(n-1) terms of A176206.
%C A340581 In other words: row n lists in nonincreasing order the terms of the first n rows of triangle A176206.
%C A340581 Conjecture: all divisors of all terms in row n are also all parts of all partitions of all positive integers <= n.
%C A340581 The conjecture is in accordance with the conjectures in A336811 and in A176206.
%C A340581 A336811 contains the most elementary conjecture about the correspondence divisors/partitions.
%C A340581 The connection with A336811 (the main sequence) is as follows: A336811 --> A176206 --> this sequence.
%e A340581 Triangle begins:
%e A340581 1;
%e A340581 2, 1, 1;
%e A340581 3, 2, 2, 1, 1, 1, 1;
%e A340581 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
%e A340581 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e A340581 ...
%e A340581 For n = 4, by definition the length of row 6 is A014153(4-1) = A014153(3) = 14, so the row 4 of triangle has 14 terms. Since every column lists the positive integers A000027 so the row 4 is [4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1].
%e A340581 Then we have that the divisors of the numbers of the 4th row are:
%e A340581 .
%e A340581 4th row of the triangle ----------> 4  3  3  2  2  2  2  1  1  1  1  1  1  1
%e A340581                                     2  1  1  1  1  1  1
%e A340581                                     1
%e A340581 .
%e A340581 There are fourteen 1's, five 2's, two 3's and one 4.
%e A340581 In total there are 14 + 5 + 2 + 1 = 22 divisors.
%e A340581 On the other hand all partitions of all positive integers <= 4 are as shown below:
%e A340581 .
%e A340581 .    Partition   Partitions    Partitions     Partitions
%e A340581 .       of 1        of 2          of 3           of 4
%e A340581 .
%e A340581 .                                             4
%e A340581 .                                             2  2
%e A340581 .                               3             3  1
%e A340581 .                   2           2  1          2  1  1
%e A340581 .        1          1  1        1  1  1       1  1  1  1
%e A340581 .
%e A340581 In these partitions there are fourteen 1's, five 2's, two 3's and one 4.
%e A340581 In total there are 14 + 5 + 2 + 1 = A284870(4) = 22 parts.
%e A340581 Finally in accordance with the conjecture we can see that all divisors of all numbers in the 4th row of the triangle are the same positive integers as all parts of all partitions of all positive integers <= 4.
%Y A340581 Cf. A000041, A000070, A027750, A014153, A176206, A284870, A336811.
%K A340581 nonn,tabf
%O A340581 1,2
%A A340581 _Omar E. Pol_, Jan 14 2021