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A047971
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Triangle of coefficients of Gaussian polynomials [ n+3,3 ].
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5
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1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 3, 4, 4, 5, 4, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 5, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 5, 7, 7, 8, 8, 8, 7, 7, 5, 4, 3, 2, 1, 1
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OFFSET
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0,8
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COMMENTS
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a(n) as illustrated is related to the following sequences: The row sum values are A001400. The column sums are A000292. The row lengths are the stuttering sequence A037915 (stutter values in A016777). The column lengths are the sequence A016777. The max values in each column are A001971. - Alford Arnold, Aug 16 2004
The Gaussian polynomial (or Gaussian binomial) [n,3]_q is an example of a q-binomial coefficient (see the link) and may be defined for n >= 3 by [n,3]_q = ([n]_q * [n-1]_q * [n-2]_q)/([1]_q * [2]_q * [3]_q), where [n]_q := q^n - 1. The first few values are: [3,3]_q = 1; [4,3]_q = 1 + q + q^2 + q^3; [5,3]_q = 1 + q + 2q^2 + 2q^3 + 2q^4 + q^5 + q^6. - Peter Bala, Sep 23 2007
The entry a(p,w), p >= 0, w = 0,1,...,3*p, of this irregular triangle is the number of nonnegative solutions of m_0 + m_1 + m_2 + m_3 = p and 1*m_1 + 2*m_2 + 3*m_3 = w. See the Hawkins reference given in A008967, p. 264, (4,7),(4.8), concerning Cayley's counting problem. N(p,3,w) there equals a(p,w). The o.g.f. has been given in the formula section by Peter Bala. See also the Cayley reference given in A008967, p. 110, 35. with m = 3, Theta = p and q = w. - Wolfdieter Lang, Dec 02 2012
The entry a(p,w) p >= 0, w = 0,1,...,3*p, of this array gives the number of partitions of w into at most p parts, each at most 3. This follows from the preceding comment with the two Diophantine equations. From Andrews, p. 33 and p. 35, a(p,w) (called there p(N,M,n) with N=p, M=3, n=w) gives also the number of partitions of w into at most 3 parts, each at most p. This is in accordance with the symmetry of the q-binomials [p+3,p] = [p+3,3]. - Wolfdieter Lang, Dec 04 2012
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REFERENCES
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G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
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LINKS
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FORMULA
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O.g.f.: 1/((1-x)(1-qx)(1-q^2x)(1-q^3x)) = 1 + x(1 + q + q^2 + q^3) + x^2(1 + q + 2q^2 + 2q^3 + 2q^4 + q^5 + q^6) + .... - Peter Bala, Sep 23 2007
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EXAMPLE
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The table a(p,w) = [q^w][p+3,3]_q starts:
p\w 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...
0: 1
1: 1 1 1 1
2: 1 1 2 2 2 1 1
3: 1 1 2 3 3 3 3 2 1 1
4: 1 1 2 3 4 4 5 4 4 3 2 1 1
5: 1 1 2 3 4 5 6 6 6 6 5 4 3 2 1 1
6: 1 1 2 3 4 5 7 7 8 8 8 7 7 5 4 3 2 1 1
Partition example: Row p=2 is 1 1 2 2 2 1 1 because there are ten solution for (m_0, m_1, m_2, m_3) of the first equation given in a comment above, namely (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1) and (0,0,1,1) which have the w = 1*m_1 + 2*m_2 + 3*m_3 values 0, 2, 4, 6, 1, 2, 3, 3, 4 and 5, respectively. Therefore there are 1, 1, 2, 2, 2, 1, 1 solutions for w = 0, 1, 2, 3, 4, 5, 6, respectively. - Wolfdieter Lang, Dec 03 2012
a(4,5) = 4 because there are 4 partitions of 5 with 1, 2, 3 or 4 parts, each being <= 3, namely all partitions of 5 excluding 5, 14 and 11111. There are also 4 partitions of 5 with 1, 2, or 3 parts, each being <= 4, namely all partitions of 5 excluding 5, 1112 and 11111. - Wolfdieter Lang, Dec 04 2012
The table may also be arranged as follows (see the Alford Arnold comment above):
1
..1
..1..1
..1..1..1
..1..2..1..1
.....2..2..1..1
.....2..3..2..1..1
.....1..3..3..2..1..1
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MATHEMATICA
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nmax = 6;
se = Series[ 1/Product[1 - q^k*x, {k, 0, 3}], {x, 0, nmax}];
row[n_] := CoefficientList[ SeriesCoefficient[se, n], q];
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CROSSREFS
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KEYWORD
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nonn,easy,nice,tabf
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AUTHOR
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STATUS
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approved
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