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%I #28 Oct 23 2023 04:31:37
%S 1,1,1,0,2,1,0,1,3,1,-1,0,3,4,1,0,-2,1,6,5,1,-1,-2,-3,4,10,6,1,0,-2,
%T -6,-3,10,15,7,1,0,-2,-6,-12,0,20,21,8,1,0,1,-6,-16,-19,9,35,28,9,1,0,
%U 0,0,-16,-35,-24,28,56,36,10,1,1,2,3,-6,-40,-65,-21,62,84,45,11,1
%N Triangle read by rows: T(n, m) gives the sum of the weights of weighted compositions of n with m parts from generalized pentagonal numbers {A001318(k)}_{k>=1}.
%C The sums of row n are given in A000041(n), for n >= 1 (number of partitions).
%C A differently signed triangle is A047265.
%C One could add a column m = 0 starting at n = 0 with T(0, 0) = 1 and T(n, 0) = 0 otherwise, by including the empty partition with no parts.
%C For the weights w of positive integer numbers n see a comment in A339885. It is w(n) = -A010815(n), for n >= 0. Also w(n) = A257028(n), for n >= 1.
%C The weight of a composition is the one of the respective partition, obtained by the product of the weights of the parts.
%C That the row sums give the number of partitions follows from the pentagonal number theorem. See also the Apr 04 2013 conjecture in A000041 by _Gary W. Adamson_, and the hint for the proof by _Joerg Arndt_. The INVERT map of A = {1, 1, 0, 0, -5, -7, ...}, with offset 1, gives the A000041(n) numbers, for n >= 0.
%C If the above mentioned column for m = 0, starting at n = 0 is added this is an ordinary convolution triangle of the Riordan type R(1, f(x)), with f(x) = -(Product_{j>=1} (1 - x^j) - 1), generating {A257628(n)_{n>=0}. See the formulae below. - _Wolfdieter Lang_, Feb 16 2021
%H Wikipedia, <a href="https://www.wikipedia.org/wiki/Pentagonal_number_theorem">Pentagonal number theorem</a>.
%F T(n, m) = Sum_{j=1..p(n,m)} w(Part(n, m, j))*M0(n, m, j), where p(n, m) = A008284(n, m), M0(n, m, j) are the multinomials from A048996, i.e., m!/Prod_{k=1..m} e(n,m,j,k)! with the exponents of the parts, and the ternary weight of the j-th partition of n with m parts Part(n,m,j), in Abramowitz-Stegun order, is defined as the product of the weights of the parts, using w(n) = -A010815(n), for n >= 1, and m = 1, 2, ..., n.
%F From _Wolfdieter Lang_, Feb 16 2021: (Start)
%F G.f. column m: G(m, x) = ( -(Product_{j>=1} (1 - x^j) - 1) )^m, for m >= 1.
%F G.f. of row polynomials R(n, x) = Sum_{m=1..n}, that is g. f. of the triangle:
%F GfT(z, x) = 1/(1 - x*G(1, z)) - 1. Riordan triangle (without m = 0 column). (End)
%e The triangle T(n, m) begins:
%e n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ... A41
%e -------------------------------------------------------------------------------
%e 1: 1 1
%e 2: 1 1 2
%e 3: 0 2 1 3
%e 4: 0 1 3 1 5
%e 5: -1 0 3 4 1 7
%e 6: 0 -2 1 6 5 1 11
%e 7: -1 -2 -3 4 10 6 1 15
%e 8: 0 -2 -6 -3 10 15 7 1 22
%e 9: 0 -2 -6 -12 0 20 21 8 1 30
%e 10: 0 1 -6 -16 -19 9 35 28 9 1 42
%e 11: 0 0 0 -16 -35 -24 28 56 36 10 1 56
%e 12: 1 2 3 -6 -40 -65 -21 62 84 45 11 1 77
%e 13: 0 2 6 8 -25 -90 -105 0 117 120 55 12 1 101
%e 14: 0 3 9 18 10 -75 -181 -148 54 200 165 66 13 1 135
%e 15: 1 0 8 28 45 -6 -189 -328 -177 162 319 220 78 14 1 176
%e 16: 0 2 6 26 75 90 -77 -419 -540 -160 352 483 286 91 15 1 231
%e 17: 0 0 0 20 80 180 140 -280 -837 -810 -44 660 702 364 105 16 1 297
%e ...
%e For instance the case n = 6: The relevant weighted partitions with parts from the pentagonal numbers and number of compositions are: m = 2: 2*(1,-5) = -2*(1,5), m = 3: 1*(2^3), m = 4: 3*(1^2,2^2), m = 5: 1*(1^4,2), m = 6: 1*(1^6). The other partitions have weight 0.
%p # Using function PMatrix from A357368. Adds a row and a column for n, m = 0.
%p PMatrix(14, proc(n) 24*n+1; if issqr(%) then sqrt(%); -(-1)^irem(iquo(%+irem(%,6),6),2) else 0 fi end); # _Peter Luschny_, Oct 06 2022
%t nmax = 12;
%t col[m_] := col[m] = (-(Product[(1-x^j), {j, 1, nmax}]-1))^m // CoefficientList[#, x]&;
%t T[n_, m_] := col[m][[n+1]];
%t Table[T[n, m], {n, 1, nmax}, {m, 1, n}] // Flatten (* _Jean-François Alcover_, Oct 23 2023 *)
%Y Cf. A000041, A008284, A010815, A047265, A257028, -A307059 (alternating row sums), A339885 (for partitions).
%K sign,tabl,easy
%O 1,5
%A _Wolfdieter Lang_, Feb 15 2021
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