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%I A260533 #37 Aug 27 2024 16:46:03
%S A260533 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,
%T A260533 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
%N 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.
%C A260533 The triangle is a refinement of Pascal's triangle A007318.
%H A260533 Johann Cigler, <a href="https://doi.org/10.48550/arXiv.2408.14094">Some elementary remarks on the powers of a partial theta function and corresponding q-analogs of the binomial coefficients</a>, arXiv:2408.14094 [math.NT], 2024.
%H A260533 Peter Luschny, <a href="https://oeis.org/wiki/User:Peter_Luschny/P-Transform">The P-transform</a>, 2016.
%H A260533 Peter Luschny, <a href="https://github.com/PeterLuschny/PartitionTransform/blob/main/PartitionTransform.ipynb">The Partition Transform</a>, A SageMath Jupyter Notebook, GitHub, 2016/2022.
%H A260533 Marko Riedel, <a href="https://math.stackexchange.com/q/4943578">Answer to Question 4943578</a>, Mathematics Stack Exchange, 2024.
%H A260533 Peter Taylor, <a href="https://mathoverflow.net/q/474504">Answer to Question 474483</a>, MathOverflow, 2024.
%F A260533 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).
%e A260533 The signed version of the triangle starts:
%e A260533 [1]
%e A260533 [-1, 1]
%e A260533 [1, -2, 1]
%e A260533 [-1, 3, -1, -2, 1]
%e A260533 [1, -4, 3, 3, -2, -2, 1]
%e A260533 [-1, 5, -6, -4, 1, 6, 3, -1, -2, -2, 1]
%e A260533 Adding adjacent coefficients with equal sign reduces the triangle to the matrix inverse of Pascal's triangle (A130595).
%e A260533 .
%e A260533 The q-polynomials defined by Cigler start:
%e A260533 [0]  1;
%e A260533 [1]  1, q;
%e A260533 [2]  1, 2*q, q^3;
%e A260533 [3]  1, 3*q, q^2+2*q^3,   q^6;
%e A260533 [4]  1, 4*q, 3*q^2+3*q^3, 2*q^4+2*q^6,     q^10;
%e A260533 [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;
%p A260533 with(combstruct): with(ListTools):
%p A260533 PartitionCoefficients := proc(n) local L, iter, p;
%p A260533 iter := iterstructs(Partition(n)): L := []:
%p A260533 while not finished(iter) do
%p A260533    p := Reverse(nextstruct(iter)):
%p A260533    L := [mul(binomial(p[j], p[j+1]), j=1..nops(p)-1), op(L)]
%p A260533 od end:
%p A260533 for n from 1 to 6 do PartitionCoefficients(n) od;
%p A260533 # Alternative, using Cigler's recurrence for the q-polynomials:
%p A260533 C := proc(n, k, q) local j;
%p A260533 if k = 0 then q^binomial(n + 1, 2) elif n = 0 then n^k else
%p A260533 add(q^binomial(j + 1, 2)*C(n - j - 1, k - 1, q), j = 0..n - k) fi end:
%p A260533 p := n -> local k; add(C(n, n - k, q)*x^k, k = 0..n):
%p A260533 row := n -> local k; seq(sort(coeff(expand(p(n)), x, k), [q], ascending), k=0..n):
%p A260533 for n from 0 to 5 do row(n) od;  # _Peter Luschny_, Aug 24 2024
%o A260533 (Sage)
%o A260533 PartitionCoeff = lambda p: mul(binomial(p[j], p[j+1]) for j in range(len(p)-1))
%o A260533 PartitionCoefficients = lambda n: [PartitionCoeff(p) for p in Partitions(n)]
%o A260533 for n in (1..7): print(PartitionCoefficients(n))
%Y A260533 Cf. A007318, A080577, A130595, A269941 (expanded form).
%K A260533 nonn,tabf
%O A260533 1,5
%A A260533 _Peter Luschny_, Jul 28 2015

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