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%I #34 Jun 01 2022 18:12:21
%S 1,4,1,8,4,2,15,8,8,3,21,15,16,12,5,33,21,30,24,20,7,41,33,42,45,40,
%T 28,11,56,41,66,63,75,56,44,15,69,56,82,99,105,105,88,60,22,87,69,112,
%U 123,165,147,165,120,88,30,99,87,138,168,205,231,231,225,176,120,42,127,99,174
%N Triangle read by rows: T(n,k) = A024916(n-k+1)*A000041(k-1), 1 <= k <= n.
%C Conjecture 1: T(n,k) is the sum of divisors of the terms that are in the k-th blocks of the first n rows of triangle A176206.
%C Conjecture 2: the sum of row n equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
%C Conjecture 3: T(n,k) is also the volume (or number of cubes) of the k-th block of a symmetric tower in which the terraces are the symmetric representation of sigma (n..1) starting from the base respectively (cf. A237270, A237593), hence the total area of the terraces is A024916(n), the same as the area of the base.
%C The levels of the terraces starting from the base are the first n terms of A000070, that is A000070(0)..A000070(n-1). Hence the differences between levels give the partition numbers A000041, that is A000041(0)..A000041(n-1).
%C This symmetric tower has the property that its volume (or total number of cubes) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
%C For another symmetric tower of the same family and whose volume equals A066186(n) see A339106 and A221529.
%C The above three conjectures are connected due to the correspondence between divisors and partitions (cf. A336811).
%e Triangle begins:
%e 1;
%e 4, 1;
%e 8, 4, 2;
%e 15, 8, 8, 3;
%e 21, 15, 16, 12, 5;
%e 33, 21, 30, 24, 20, 7;
%e 41, 33, 42, 45, 40, 28, 11;
%e 56, 41, 66, 63, 75, 56, 44, 15;
%e 69, 56, 82, 99, 105, 105, 88, 60, 22;
%e 87, 69, 112, 123, 165, 147, 165, 120, 88, 30;
%e 99, 87, 138, 168, 205, 231, 231, 225, 176, 120, 42;
%e ...
%e For n = 6 the calculation of every term of row 6 is as follows:
%e --------------------------
%e k A000041 T(6,k)
%e 1 1 * 33 = 33
%e 2 1 * 21 = 21
%e 3 2 * 15 = 30
%e 4 3 * 8 = 24
%e 5 5 * 4 = 20
%e 6 7 * 1 = 7
%e . A024916
%e --------------------------
%e The sum of row 6 is 33 + 21 + 30 + 24 + 20 + 7 = 135, equaling A182738(6).
%Y Columns 1 and 2 give A024916.
%Y Column 3 gives A327329.
%Y Leading diagonal gives A000041.
%Y Row sums give A182738.
%Y Cf. A000070, A066186, A176206, A221529, A221531, A237270, A237593, A336811, A336812, A338156, A339106, A340035, A340424, A340425, A340426, A340524, A340526.
%K nonn,tabl
%O 1,2
%A _Omar E. Pol_, Jan 10 2021
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