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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 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.)