A092905 Triangle, read by rows, such that the partial sums of the n-th row form the n-th diagonal, for n>=0, where each row begins with 1.

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 6, 4, 2, 1, 1, 6, 9, 7, 4, 2, 1, 1, 7, 12, 11, 7, 4, 2, 1, 1, 8, 16, 16, 12, 7, 4, 2, 1, 1, 9, 20, 23, 18, 12, 7, 4, 2, 1, 1, 10, 25, 31, 27, 19, 12, 7, 4, 2, 1, 1, 11, 30, 41, 38, 29, 19, 12, 7, 4, 2, 1, 1, 12, 36, 53, 53, 42, 30, 19, 12, 7, 4, 2, 1

Offset: 0

Author

Paul D. Hanna, Mar 12 2004

Keywords

• nonn
• tabl

Comments

Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070).

From Alford Arnold, Feb 07 2010: (Start)

The table can also be generated by summing sequences embedded within Table A008284

For example,

1 1 1 1 ... yields 1 2 3 4 ...

1 1 2 2 3 3 ... yields 1 2 4 6 9 12 ...

1 1 2 3 4 5 7 ... yields 1 2 4 7 11 16 ...

(End)

T(n,k) is also count of all 'replacable' cells in the (Ferrers plots of) the partitions on n in exactly k parts. [Wouter Meeussen, Sep 16 2010]

From Wolfdieter Lang, Dec 03 2012: (Start)

The triangle entry T(n,k) is obtained from triangle A072233 by summing the entries of column k up to n (see the partial sum type o.g.f. given by Vladeta Jovovic in the formula section).

Therefore, the o.g.f. for the sequence in column k is x^k/((1-x)* product(1-x^j,j=1..k)).

The triangle with entry a(n,m) = T(n-1,m-1), n >= 1, m = 1, ..., n, is obtained from the partition array A103921 when in row n all entries belonging to part number m are summed (a conjecture). (End)

Examples

			The fourth row (n=3) is {1,3,2,1} and the fourth diagonal is the partial sums of the fourth row: {1,4,6,7,7,7,7,7,...}.
The triangle T(n,k) begins:
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12  ...
0   1
1   1  1
2   1  2  1
3   1  3  2  1
4   1  4  4  2  1
5   1  5  6  4  2  1
6   1  6  9  7  4  2  1
7   1  7 12 11  7  4  2  1
8   1  8 16 16 12  7  4  2  1
9   1  9 20 23 18 12  7  4  2  1
10  1 10 25 31 27 19 12  7  4  2  1
11  1 11 30 41 38 29 19 12  7  4  2  1
12  1 12 36 53 53 42 30 19 12  7  4  2  1
... Reformatted by _Wolfdieter Lang_, Dec 03 2012
T(5,3)=4 because the partitions of 5 in exactly 3 parts are 221 and 311, and they give rise to partitions of 4 in four ways: 221->22 and 211, 311->211 and 31, since both their Ferrers plots have 2 'mobile cells' each. [_Wouter Meeussen_, Sep 16 2010]
T(5,3) = a(6,4) = 4 because the partitions of 6 with 4 parts are 1113 and 1122, with the number of distinct parts 2 and 2, respectively, summing to 4 (see the array A103921). An example for the conjecture given as comment above. - _Wolfdieter Lang_, Dec 03 2012
		

Links

• V. V. Kruchinin, The number of partitions of a natural number n into parts each of which is not less than m, Math. Notes 86 (4) (2009) 505-509
• R. J. Mathar, Size of the set of residues of integer powers of fixed exponent, (2017), Table 11.

Crossrefs

Antidiagonal sums form the partition numbers (A000041).

Cf. A000070.

Cf. A008284. [Alford Arnold, Feb 07 2010]

Columns: A087811, A000601, A002621, A002622, A288341 - A288345.

Programs

Maple

T(n,k)=if(n
				

Mathematica

(*Needs["DiscreteMath`Combinatorica`"]; partitionexact[n_, m_] := TransposePartition /@ (Prepend[ #1, m] & ) /@ Partitions[n - m, m] *); mobile[p_?PartitionQ]:=1+Count[Drop[p,-1]-Rest[p],?Positive]; Table[Tr[mobile/@partitionexact[n,k]],{n,12},{k,n}] (* _Wouter Meeussen, Sep 16 2010 *)

Formula

T(n, k) = sum_{j=0..k} T(n-k, j), with T(n, 0) = 1 for all n>=0. A000070(n) = sum_{k=0..n} T(n, k).

O.g.f.: (1/(1-y))*(1/Product(1-x*y^k, k=1..infinity)). - Vladeta Jovovic, Jan 29 2005

Extensions

Several corrections by Wolfdieter Lang, Dec 03 2012