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A340061 Irregular triangle read by rows T(n,k) in which row n lists n blocks, where the m-th block consists of A000041(n-m) copies of m, with n >= 1 and m >= 1.

%I A340061 #80 Sep 03 2023 11:16:02
%S A340061 1,1,2,1,1,2,3,1,1,1,2,2,3,4,1,1,1,1,1,2,2,2,3,3,4,5,1,1,1,1,1,1,1,2,
%T A340061 2,2,2,2,3,3,3,4,4,5,6,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,
%U A340061 4,4,4,5,5,6,7,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2
%N A340061 Irregular triangle read by rows T(n,k) in which row n lists n blocks, where the m-th block consists of A000041(n-m) copies of m, with n >= 1 and m >= 1.
%C A340061 Conjecture: all divisors of all terms of row n are also all parts of all partitions of n.
%C A340061 The conjecture gives a correspondence between divisors and partitions (see example).
%C A340061 It is conjectured that every section of the set of partitions of n has essentially the same correspondence. For more information see A336811.
%H A340061 Paolo Xausa, <a href="/A340061/b340061.txt">Table of n, a(n) for n = 1..10980</a> (rows 1..21 of the triangle, flattened)
%e A340061 Triangle begins:
%e A340061   1;
%e A340061   1, 2;
%e A340061   1, 1, 2, 3;
%e A340061   1, 1, 1, 2, 2, 3, 4;
%e A340061   1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5;
%e A340061   1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6;
%e A340061   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, ...
%e A340061   ...
%e A340061 For n = 6 the 6th row of the triangle consists of:
%e A340061   p(5) = 7 copies of 1, that is, [1, 1, 1, 1, 1, 1, 1],
%e A340061   p(4) = 5 copies of 2, that is, [2, 2, 2, 2, 2],
%e A340061   p(3) = 3 copies of 3, that is, [3, 3, 3],
%e A340061   p(2) = 2 copies of 4, that is, [4, 4],
%e A340061   p(1) = 1 copy   of 5, that is, [5],
%e A340061   p(0) = 1 copy   of 6, that is, [6],
%e A340061 where p(j) is the j-th partition number A000041(j).
%e A340061 About the conjecture we have that the divisors of the terms of the 6th row are:
%e A340061                                                                      1
%e A340061                                                             1, 1,    2
%e A340061                                     1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 3
%e A340061   6th row -->  1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6
%e A340061 There are nineteen 1's, eight 2's, four 3's, two 4's, one 5 and one 6.
%e A340061 In total there are 19 + 8 + 4 + 2 + 1 + 1 = 35 divisors.
%e A340061 On the other hand the partitions of 6 are:
%e A340061       Diagram          Parts
%e A340061     _ _ _ _ _ _
%e A340061    |_ _ _      |       6
%e A340061    |_ _ _|_    |       3 3
%e A340061    |_ _    |   |       4 2
%e A340061    |_ _|_ _|_  |       2 2 2
%e A340061    |_ _ _    | |       5 1
%e A340061    |_ _ _|_  | |       3 2 1
%e A340061    |_ _    | | |       4 1 1
%e A340061    |_ _|_  | | |       2 2 1 1
%e A340061    |_ _  | | | |       3 1 1 1
%e A340061    |_  | | | | |       2 1 1 1 1
%e A340061    |_|_|_|_|_|_|       1 1 1 1 1 1
%e A340061 There are nineteen 1's, eight 2's, four 3's, two 4's, one 5 and one 6, as shown also the 6th row of A066633.
%e A340061 In total there are 19 + 8 + 4 + 2 + 1 + 1 = A006128(6) = 35 parts.
%e A340061 In accordance with the conjecture we can see that all divisors of all terms of the 6th row of triangle are the same positive integers as all parts of all partitions of 6.
%t A340061 A340061row[n_]:=Flatten[Table[ConstantArray[m,PartitionsP[n-m]],{m,n}]];Array[A340061row,10] (* _Paolo Xausa_, Sep 01 2023 *)
%Y A340061 Mirror of A176206.
%Y A340061 Row sums give A014153.
%Y A340061 Row n has length A000070(n-1).
%Y A340061 Right border gives A000027.
%Y A340061 Cf. A000041, A006128, A066633, A027750, A066186, A176206, A336811 (section).
%K A340061 nonn,tabf
%O A340061 1,3
%A A340061 _Omar E. Pol_, Dec 28 2020