A260533 Table of partition coefficients read by rows. The coefficient of a partition p is Product_{j=1..length(p)-1} C(p[j], p[j+1]). Row n lists the coefficients of the partitions of n in the ordering A080577, for n>=1.

1, 1, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 4, 3, 3, 2, 2, 1, 1, 5, 6, 4, 1, 6, 3, 1, 2, 2, 1, 1, 6, 10, 5, 4, 12, 4, 3, 3, 6, 3, 2, 2, 2, 1, 1, 7, 15, 6, 10, 20, 5, 1, 12, 6, 12, 4, 3, 3, 6, 6, 3, 1, 2, 2, 2, 1

Offset

1,5

Comments

The triangle is a refinement of Pascal's triangle A007318.

Links

Table of n, a(n) for n=1..66.

Formula

Let P = Partitions(n, k) denote the set of partitions p of n with largest part k. Then Sum_{p in P} PartitionCoefficient(p) = binomial(n-1,k-1) for n>=0 and k>=0 (assuming binomial(-1,-1) = 1).

Example

The signed version of the triangle starts:
[1]
[-1, 1]
[1, -2, 1]
[-1, 3, -1, -2, 1]
[1, -4, 3, 3, -2, -2, 1]
[-1, 5, -6, -4, 1, 6, 3, -1, -2, -2, 1]
Adding adjacent coefficients with equal sign reduces the triangle to the matrix inverse of Pascal's triangle (A130595).

Maple

with(combstruct): with(ListTools):
PartitionCoefficients := proc(n) local L, iter, p;
iter := iterstructs(Partition(n)): L := []:
while not finished(iter) do
   p := Reverse(nextstruct(iter)):
   L := [mul(binomial(p[j], p[j+1]), j=1..nops(p)-1), op(L)]
od end:
for n from 1 to 6 do PartitionCoefficients(n) od;

Prog

(Sage)
PartitionCoeff = lambda p: mul(binomial(p[j], p[j+1]) for j in range(len(p)-1))
PartitionCoefficients = lambda n: [PartitionCoeff(p) for p in Partitions(n)]
for n in (1..7): print(PartitionCoefficients(n))

Crossrefs

Cf. A007318, A080577, A130595.

Sequence in context: A360056 A244316 A076259 * A107359 A307641 A342255

Adjacent sequences: A260530 A260531 A260532 * A260534 A260535 A260536

Keyword

nonn,tabf

Author

Peter Luschny, Jul 28 2015

Status

approved