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

0,5

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)

Links

Table of n, a(n) for n=0..90.

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.

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

Example

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

Maple

T(n, k)=if(n<k|k<0, 0, if(n==k|k==0, 1, sum(j=0, min(k, n-k), T(n-k, j))))

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 *)

Crossrefs

Antidiagonal sums form the partition numbers (A000041).

Cf. A000070.

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

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

Sequence in context: A343658 A194005 A055794 * A052509 A172119 A228125

Adjacent sequences: A092902 A092903 A092904 * A092906 A092907 A092908

Keyword

nonn,tabl

Author

Paul D. Hanna, Mar 12 2004

Extensions

Several corrections by Wolfdieter Lang, Dec 03 2012

Status

approved