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%I #32 Feb 02 2022 09:26:43
%S 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,
%T 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,
%U 1,1
%N Triangle of coefficients of Gaussian polynomials [ n+3,3 ].
%C 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
%C 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
%C 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
%C 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
%D G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/q-BinomialCoefficient.html">q-Binomial Coefficient</a>.
%F 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
%e The table a(p,w) = [q^w][p+3,3]_q starts:
%e p\w 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...
%e 0: 1
%e 1: 1 1 1 1
%e 2: 1 1 2 2 2 1 1
%e 3: 1 1 2 3 3 3 3 2 1 1
%e 4: 1 1 2 3 4 4 5 4 4 3 2 1 1
%e 5: 1 1 2 3 4 5 6 6 6 6 5 4 3 2 1 1
%e 6: 1 1 2 3 4 5 7 7 8 8 8 7 7 5 4 3 2 1 1
%e ... Reformatted and extended by _Wolfdieter Lang_, Dec 04 2012
%e 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
%e 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
%e The table may also be arranged as follows (see the Alford Arnold comment above):
%e 1
%e ..1
%e ..1..1
%e ..1..1..1
%e ..1..2..1..1
%e .....2..2..1..1
%e .....2..3..2..1..1
%e .....1..3..3..2..1..1
%t nmax = 6;
%t se = Series[ 1/Product[1 - q^k*x, {k, 0, 3}], {x, 0, nmax}];
%t row[n_] := CoefficientList[ SeriesCoefficient[se, n], q];
%t Flatten[ Table[ row[n], {n, 0, nmax}]] (* _Jean-François Alcover_, Dec 19 2011 *)
%Y Cf. A008967.
%Y Cf. A001400.
%K nonn,easy,nice,tabf
%O 0,8
%A _N. J. A. Sloane_
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