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%I #48 Jan 10 2021 11:21:36
%S 1,3,1,4,3,2,7,4,6,3,6,7,8,9,5,12,6,14,12,15,7,8,12,12,21,20,21,11,15,
%T 8,24,18,35,28,33,15,13,15,16,36,30,49,44,45,22,18,13,30,24,60,42,77,
%U 60,66,30,12,18,26,45,40,84,66,105,88,90,42,28,12,36,39,75,56,132,90,154,120,126,56
%N Triangle read by rows: T(n,k) = A000203(n-k+1)*A000041(k-1), n >= 1, 1 <= k <= n.
%C Conjecture 1: T(n,k) is the sum of all divisors of all (n - k + 1)'s in the n-th row of triangle A176206, assuming that A176206 has offset 1. The same for the triangle A340061.
%C Conjecture 2: the sum of row n equals A066186(n), the sum of all parts of all partitions of n.
%F T(n,k) = sigma(n-k+1)*p(k-1), n >= 1, 1 <= k <= n.
%e Triangle begins:
%e 1;
%e 3, 1;
%e 4, 3, 2;
%e 7, 4, 6, 3;
%e 6, 7, 8, 9, 5;
%e 12, 6, 14, 12, 15, 7;
%e 8, 12, 12, 21, 20, 21, 11;
%e 15, 8, 24, 18, 35, 28, 33, 15;
%e 13, 15, 16, 36, 30, 49, 44, 45, 22;
%e 18, 13, 30, 24, 60, 42, 77, 60, 66, 30;
%e 12, 18, 26, 45, 40, 84, 66, 105, 88, 90, 42;
%e 28, 12, 36, 39, 75, 56, 132, 90, 154, 120, 126, 56;
%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 * 12 = 12
%e 2 1 * 6 = 6
%e 3 2 * 7 = 14
%e 4 3 * 4 = 12
%e 5 5 * 3 = 15
%e 6 7 * 1 = 7
%e . A000203
%e -------------------------
%e The sum of row 6 is 12 + 6 + 14 + 12 + 15 + 7 = 66, equaling A066186(6).
%t T[n_, k_] := DivisorSigma[1, n - k + 1] * PartitionsP[k - 1]; Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Amiram Eldar_, Jan 08 2021 *)
%o (PARI) T(n, k) = sigma(n-k+1)*numbpart(k-1); \\ _Michel Marcus_, Jan 08 2021
%Y Mirror of A221529.
%Y Row sums give A066186 (conjectured).
%Y Main diagonal gives A000041.
%Y Columns 1 and 2 give A000203.
%Y Column 3 gives A074400.
%Y Column 4 gives A272027.
%Y Column 5 gives A274535.
%Y Column 6 gives A319527.
%Y Cf. A176206, A340061.
%K nonn,tabl
%O 1,2
%A _Omar E. Pol_, Nov 23 2020
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