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A182700 Triangle T(n,k) = n*A000041(n-k), 0<=k<=n, read by rows.

%I A182700 #24 Jun 23 2020 19:03:52
%S A182700 0,1,1,4,2,2,9,6,3,3,20,12,8,4,4,35,25,15,10,5,5,66,42,30,18,12,6,6,
%T A182700 105,77,49,35,21,14,7,7,176,120,88,56,40,24,16,8,8,270,198,135,99,63,
%U A182700 45,27,18,9,9,420,300,220,150,110,70,50,30,20,10,10,616,462,330,242,165,121,77,55,33
%N A182700 Triangle T(n,k) = n*A000041(n-k), 0<=k<=n, read by rows.
%C A182700 T(n,k) is the sum of the parts of all partitions of n that contain k as a part, assuming that all partitions of n have 0 as a part: Thus, column 0 gives the sum of the parts of all partitions of n.
%C A182700 By definition all entries in row n>0 are divisible by n.
%C A182700 Row sums are 0, 2, 8, 21, 48, 95, 180, 315, 536, 873, 1390, 2145,...
%C A182700 The partitions of n+k that contain k as a part can be obtained by adding k to every partition of n assuming that all partitions of n have 0 as a part.
%C A182700 For example, the partitions of 6+k that contain k as a part are
%C A182700 k + 6
%C A182700 k + 3 + 3
%C A182700 k + 4 + 2
%C A182700 k + 2 + 2 + 2
%C A182700 k + 5 + 1
%C A182700 k + 3 + 2 + 1
%C A182700 k + 4 + 1 + 1
%C A182700 k + 2 + 2 + 1 + 1
%C A182700 k + 3 + 1 + 1 + 1
%C A182700 k + 2 + 1 + 1 + 1 + 1
%C A182700 k + 1 + 1 + 1 + 1 + 1 + 1
%C A182700 The partition number A000041(n) is also the number of partitions of m*(n+k) into parts divisible by m and that contain m*k as a part, with k>=0, m>=1, n>=0 and assuming that all partitions of n have 0 as a part.
%H A182700 Robert Price, <a href="/A182700/b182700.txt">Table of n, a(n) for n = 0..5150</a> (First 100 rows)
%F A182700 T(n,0) = A066186(n).
%F A182700 T(n,k) = A182701(n,k), n>=1 and k>=1.
%F A182700 T(n,n) = n = min { T(n,k); 0<=k<=n }.
%e A182700 For n=7 and k=4 there are 3 partitions of 7 that contain 4 as a part. These partitions are (4+3)=7, (4+2+1)=7 and (4+1+1+1)=7. The sum is 7+7+7 = 7*3 = 21. By other way, the partition number of 7-4 is A000041(3) = p(3)=3, then 7*3 = 21, so T(7,4) = 21.
%e A182700 Triangle begins with row n=0 and columns 0<=k<=n :
%e A182700 0,
%e A182700 1, 1,
%e A182700 4, 2, 2,
%e A182700 9, 6, 3, 3,
%e A182700 20,12,8, 4, 4,
%e A182700 35,25,15,10,5, 5,
%e A182700 66,42,30,18,12,6, 6
%p A182700 A182700 := proc(n,k) n*combinat[numbpart](n-k) ; end proc:
%p A182700 seq(seq(A182700(n,k),k=0..n),n=0..15) ;
%t A182700 Table[n*PartitionsP[n-k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Robert Price_, Jun 23 2020 *)
%o A182700 (PARI) A182700(n,k) = n*numbpart(n-k)
%Y A182700 Cf. A000041, A027293, A135010, A138121.
%Y A182700 Two triangles that are essentially the same as this are A027293 and A140207. - _N. J. A. Sloane_, Nov 28 2010
%Y A182700 Row sums give A182704.
%K A182700 nonn,tabl
%O A182700 0,4
%A A182700 _Omar E. Pol_, Nov 27 2010