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%I #47 Sep 04 2023 06:07:12
%S 1,1,2,2,2,2,3,4,2,3,5,6,4,3,2,7,10,6,6,2,4,11,14,10,9,4,4,2,15,22,14,
%T 15,6,8,2,4,22,30,22,21,10,12,4,4,3,30,44,30,33,14,20,6,8,3,4,42,60,
%U 44,45,22,28,10,12,6,4,2,56,84,60,66,30,44,14,20,9,8,2,6
%N Triangle read by rows: T(n,k) = A000005(k)*A000041(n-k).
%C T(n,k) is the number of partitions of n that contain k as a part multiplied by the number of divisors of k.
%C It appears that T(n,k) is also the total number of appearances of k in the last k sections of the set of partitions of n multiplied by the number of divisors of k.
%C T(n,k) is also the number of partitions of k into equal parts multiplied by the number of ones in the j-th section of the set of partitions of n, where j = (n - k + 1).
%C For another version see A245095. - _Omar E. Pol_, Jul 15 2014
%H Paolo Xausa, <a href="/A221530/b221530.txt">Table of n, a(n) for n = 1..11325</a> (rows 1..150 of the triangle, flattened)
%F T(n,k) = d(k)*p(n-k) = A000005(k)*A027293(n,k).
%e For n = 6:
%e -------------------------
%e k A000005 T(6,k)
%e 1 1 * 7 = 7
%e 2 2 * 5 = 10
%e 3 2 * 3 = 6
%e 4 3 * 2 = 6
%e 5 2 * 1 = 2
%e 6 4 * 1 = 4
%e . A000041
%e -------------------------
%e So row 6 is [7, 10, 6, 6, 4, 2]. Note that the sum of row 6 is 7+10+6+6+2+4 = 35 equals A006128(6).
%e .
%e Triangle begins:
%e 1;
%e 1, 2;
%e 2, 2, 2;
%e 3, 4, 2, 3;
%e 5, 6, 4, 3, 2;
%e 7, 10, 6, 6, 2, 4;
%e 11, 14, 10, 9, 4, 4, 2;
%e 15, 22, 14, 15, 6, 8, 2, 4;
%e 22, 30, 22, 21, 10, 12, 4, 4, 3;
%e 30, 44, 30, 33, 14, 20, 6, 8, 3, 4;
%e 42, 60, 44, 45, 22, 28, 10, 12, 6, 4, 2;
%e 56, 84, 60, 66, 30, 44, 14, 20, 9, 8, 2, 6;
%e ...
%t A221530row[n_]:=DivisorSigma[0,Range[n]]PartitionsP[n-Range[n]];Array[A221530row,10] (* _Paolo Xausa_, Sep 04 2023 *)
%o (PARI) row(n) = vector(n, i, numdiv(i)*numbpart(n-i)); \\ _Michel Marcus_, Jul 18 2014
%Y Similar to A221529.
%Y Columns 1-2: A000041, A139582. Leading diagonals 1-3: A000005, A000005, A062011. Row sums give A006128.
%Y Cf. A027293, A135010, A138137, A182703, A245095, A245099.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Jan 19 2013
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