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
Examples
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). . The q-polynomials defined by Cigler start: [0] 1; [1] 1, q; [2] 1, 2*q, q^3; [3] 1, 3*q, q^2+2*q^3, q^6; [4] 1, 4*q, 3*q^2+3*q^3, 2*q^4+2*q^6, q^10; [5] 1, 5*q, 6*q^2+4*q^3, q^3+6*q^4+3*q^6, q^6+2*q^7+2*q^10, q^15;
Links
- Johann Cigler, Some elementary remarks on the powers of a partial theta function and corresponding q-analogs of the binomial coefficients, arXiv:2408.14094 [math.NT], 2024.
- Peter Luschny, The P-transform, 2016.
- Peter Luschny, The Partition Transform, A SageMath Jupyter Notebook, GitHub, 2016/2022.
- Marko Riedel, Answer to Question 4943578, Mathematics Stack Exchange, 2024.
- Peter Taylor, Answer to Question 474483, MathOverflow, 2024.
Programs
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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; # Alternative, using Cigler's recurrence for the q-polynomials: C := proc(n, k, q) local j; if k = 0 then q^binomial(n + 1, 2) elif n = 0 then n^k else add(q^binomial(j + 1, 2)*C(n - j - 1, k - 1, q), j = 0..n - k) fi end: p := n -> local k; add(C(n, n - k, q)*x^k, k = 0..n): row := n -> local k; seq(sort(coeff(expand(p(n)), x, k), [q], ascending), k=0..n): for n from 0 to 5 do row(n) od; # Peter Luschny, Aug 24 2024
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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))
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).
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