A182700 - OEIS

%I #24 Jun 23 2020 19:03:52

%S 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 105,77,49,35,21,14,7,7,176,120,88,56,40,24,16,8,8,270,198,135,99,63,

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

%C 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 By definition all entries in row n>0 are divisible by n.

%C Row sums are 0, 2, 8, 21, 48, 95, 180, 315, 536, 873, 1390, 2145,...

%C 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 For example, the partitions of 6+k that contain k as a part are

%C k + 6

%C k + 3 + 3

%C k + 4 + 2

%C k + 2 + 2 + 2

%C k + 5 + 1

%C k + 3 + 2 + 1

%C k + 4 + 1 + 1

%C k + 2 + 2 + 1 + 1

%C k + 3 + 1 + 1 + 1

%C k + 2 + 1 + 1 + 1 + 1

%C k + 1 + 1 + 1 + 1 + 1 + 1

%C 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 Robert Price, <a href="/A182700/b182700.txt">Table of n, a(n) for n = 0..5150</a> (First 100 rows)

%F T(n,0) = A066186(n).

%F T(n,k) = A182701(n,k), n>=1 and k>=1.

%F T(n,n) = n = min { T(n,k); 0<=k<=n }.

%e 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 Triangle begins with row n=0 and columns 0<=k<=n :

%e 0,

%e 1, 1,

%e 4, 2, 2,

%e 9, 6, 3, 3,

%e 20,12,8, 4, 4,

%e 35,25,15,10,5, 5,

%e 66,42,30,18,12,6, 6

%p A182700 := proc(n,k) n*combinat[numbpart](n-k) ; end proc:

%p seq(seq(A182700(n,k),k=0..n),n=0..15) ;

%t Table[n*PartitionsP[n-k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Robert Price_, Jun 23 2020 *)

%o (PARI) A182700(n,k) = n*numbpart(n-k)

%Y Cf. A000041, A027293, A135010, A138121.

%Y Two triangles that are essentially the same as this are A027293 and A140207. - _N. J. A. Sloane_, Nov 28 2010

%Y Row sums give A182704.

%K nonn,tabl

%O 0,4

%A _Omar E. Pol_, Nov 27 2010