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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. 0
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 (list; graph; refs; listen; history; text; internal format)
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: A228531 A244316 A076259 * A107359 A307641 A112377

Adjacent sequences:  A260530 A260531 A260532 * A260534 A260535 A260536

KEYWORD

nonn,tabf

AUTHOR

Peter Luschny, Jul 28 2015

STATUS

approved

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Last modified July 16 00:52 EDT 2019. Contains 325061 sequences. (Running on oeis4.)