A047971 Triangle of coefficients of Gaussian polynomials [ n+3,3 ].
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
Offset: 0
Examples
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 ... Reformatted and extended by _Wolfdieter Lang_, Dec 04 2012 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
References
- G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
Links
- Eric Weisstein's World of Mathematics, q-Binomial Coefficient.
Programs
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Mathematica
nmax = 6; se = Series[ 1/Product[1 - q^k*x, {k, 0, 3}], {x, 0, nmax}]; row[n_] := CoefficientList[ SeriesCoefficient[se, n], q]; Flatten[ Table[ row[n], {n, 0, nmax}]] (* Jean-François Alcover, Dec 19 2011 *)
Formula
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
Comments