%I
%S 1,1,3,2,3,4,3,6,4,7,5,9,8,7,6,7,15,12,14,6,12,11,21,20,21,12,12,8,15,
%T 33,28,35,18,24,8,15,22,45,44,49,30,36,16,15,13,30,66,60,77,42,60,24,
%U 30,13,18,42,90,88,105,66,84,40,45,26,18,12,56,126,120,154,90,132,56,75,39,36,12,28
%N Triangle read by rows: T(n,k) = A000203(k)*A000041(nk).
%C Row sums give A066186.
%C Column 1 is A000041.
%C Leading diagonals 12: A000203, A000203.
%C T(n,k) is the number of partitions of n that contain k as a part multiplied by the sum of divisors of k.
%C It appears that T(n,k) is also the number of appearances of k in the last k section of the set of partitions of n multiplied by the sum of divisors of k.
%C T(n,k) is also the total number of parts in all partitions of k into equal parts multiplied by the number of ones in the jth section of the set of partitions of n, where j = (n  k + 1).
%C Since A000203(k) has a symmetric representation then both T(n,k) and the partial sums of row n can be represented by symmetric polycubes  for more information see A237593 and A237270. For another version see A245099.  _Omar E. Pol_, Jul 15 2014
%F T(n,k) = sigma(k)*p(nk) = A000203(k)*A027293(n,k).
%e For n = 6:
%e 
%e k A000203 T(6,k)
%e 1 1 * 7 = 7
%e 2 3 * 5 = 15
%e 3 4 * 3 = 12
%e 4 7 * 2 = 14
%e 5 6 * 1 = 6
%e 6 12 * 1 = 12
%e . A000041
%e 
%e So row 6 is [7, 15, 12, 14, 6, 12]. Note that the sum of row 6 is 7+15+12+14+6+12 = 66 equals A066186(6) = 6*p(6) = 6*11 = 66.
%e Triangle begins:
%e 1;
%e 1, 3;
%e 2, 3, 4;
%e 3, 6, 4, 7;
%e 5, 9, 8, 7, 6;
%e 7, 15, 12, 14, 6, 12;
%e 11, 21, 20, 21, 12, 12, 8;
%e 15, 33, 28, 35, 18, 24, 8, 15;
%e 22, 45, 44, 49, 30, 36, 16, 15, 13;
%e 30, 66, 60, 77, 42, 60, 24, 30, 13, 18;
%e 42, 90, 88, 105, 66, 84, 40, 45, 26, 18, 12;
%e 56, 126, 120, 154, 90, 132, 56, 75, 39, 36, 12, 28;
%o (PARI) T(n,k)=sigma(k)*numbpart(nk) \\ _Charles R Greathouse IV_, Feb 19 2013
%Y Cf. A000041, A000203, A027293, A066186, A135010, A138137, A182703, A221530, A245095, A245099.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Jan 20 2013
