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A341418
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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}.
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4
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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, -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, 0, 0, -16, -35, -24, 28, 56, 36, 10, 1, 1, 2, 3, -6, -40, -65, -21, 62, 84, 45, 11, 1
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OFFSET
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1,5
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COMMENTS
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The sums of row n are given in A000041(n), for n >= 1 (number of partitions).
A differently signed triangle is A047265.
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.
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.
The weight of a composition is the one of the respective partition, obtained by the product of the weights of the parts.
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.
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
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LINKS
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FORMULA
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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.
G.f. column m: G(m, x) = ( -(Product_{j>=1} (1 - x^j) - 1) )^m, for m >= 1.
G.f. of row polynomials R(n, x) = Sum_{m=1..n}, that is g. f. of the triangle:
GfT(z, x) = 1/(1 - x*G(1, z)) - 1. Riordan triangle (without m = 0 column). (End)
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EXAMPLE
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The triangle T(n, m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ... A41
-------------------------------------------------------------------------------
1: 1 1
2: 1 1 2
3: 0 2 1 3
4: 0 1 3 1 5
5: -1 0 3 4 1 7
6: 0 -2 1 6 5 1 11
7: -1 -2 -3 4 10 6 1 15
8: 0 -2 -6 -3 10 15 7 1 22
9: 0 -2 -6 -12 0 20 21 8 1 30
10: 0 1 -6 -16 -19 9 35 28 9 1 42
11: 0 0 0 -16 -35 -24 28 56 36 10 1 56
12: 1 2 3 -6 -40 -65 -21 62 84 45 11 1 77
13: 0 2 6 8 -25 -90 -105 0 117 120 55 12 1 101
14: 0 3 9 18 10 -75 -181 -148 54 200 165 66 13 1 135
15: 1 0 8 28 45 -6 -189 -328 -177 162 319 220 78 14 1 176
16: 0 2 6 26 75 90 -77 -419 -540 -160 352 483 286 91 15 1 231
17: 0 0 0 20 80 180 140 -280 -837 -810 -44 660 702 364 105 16 1 297
...
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.
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MAPLE
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# Using function PMatrix from A357368. Adds a row and a column for n, m = 0.
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
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MATHEMATICA
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nmax = 12;
col[m_] := col[m] = (-(Product[(1-x^j), {j, 1, nmax}]-1))^m // CoefficientList[#, x]&;
T[n_, m_] := col[m][[n+1]];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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