A219237 Coefficient of Gauss polynomials [n+4,4]_q (q-binomials).

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 4, 5, 4, 4, 3, 2, 1, 1, 1, 1, 2, 3, 5, 5, 7, 7, 8, 7, 7, 5, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 6, 8, 9, 11, 11, 12, 11, 11, 9, 8, 6, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 6, 9, 10, 13, 14, 16, 16, 18, 16, 16, 14, 13, 10, 9, 6, 5, 3, 2, 1, 1

Offset: 0

Author

Wolfdieter Lang, Dec 04 2012

Keywords

• nonn
• easy
• tabf

Comments

The length of row n of this table is 4*n + 1 = A016813(n).

The sum of row n is binomial(n+4,4) = A000332(n+4), n>= 0.

The Gauss polynomial [n+4,4]q := [n+4]_q/([n]_q*[4]_q), with [n]_q = Product{j=1..n} (1-q^j) = (q;q)n (in q-shifted factorials notation), n>=0. [n+4,4]_q = (Product{j=(n+1)..(n+4)} (1-q^j))/(Product_{j=1..4} (1-q^j)). This is a polynomial in q (of degree 4*n) because it is the o.g.f. of the numbers p(n,4,k), the number of partitions of k into at most 4 parts, each <= n (see Andrews, p. 33 and 35). p(n,4,k) is also the number of partitions of k into at most n parts, each <= 4, due to the symmetry property [n+4,4]q = [n+4,n]_q (Andrews, (3,3,2), p.35). With the latter interpretation p(n,4,k) is the number of solutions of the two Diophantine equations Sum{j=1..4} j*m(j) = k and Sum_{j=0..m} m(j) = n, i.e. Sum_{j=1..m} m(j) = n - m(0), with 0 <= m(j) <= n. Therefore p(n,4,k) = [q^k] [x^n] G(4;x,q) with o.g.f. G(4;x,q) = 1/Product_{j=0..4} (1-x*q^j). Here we will call p(n,4,k) = T(n,k), n >= 0, 0 <= k <= 4*n.

See the comments in A008967 concerning a counting problem of Cayley (there m = 4, Theta = n and q = k), described also in the Hawkins reference (N(p->n,4,w->k) = T(n,k)) given there.

Examples

			The triangle T(n,k) begins:
  n\k 0  1  2  3  4  5  6   7   8   9  10  11  12  13  14  15  16 ...
  0:  1
  1:  1  1  1  1  1
  2:  1  1  2  2  3  2  2   1   1
  3:  1  1  2  3  4  4  5   4   4   3   2   1   1
  4:  1  1  2  3  5  5  7   7   8   7   7   5   5   3   2   1   1
  5:  1  1  2  3  5  6  8   9  11  11  12  11  11   9   8   6   5   3  2  1  1
  6:  1  1  2  3  5  6  9  10  13  14  16  16  18  16  16  14  13  10  9  6  5  3  2  1 1
Partition interpretation: T(3,5) = 4 because there are 4 partitions of 5 into at most 4 parts, each <= 3, namely 23, 113, 122 and 1112. here are also 4 partitions of 5 into at most 3 parts, each <= 4, namely 14, 23, 113 and 122. Note the conjugacy of the partitions 1112 and 14.
The 4 solutions of the two Diophantine equations given in a comment, with k=5 and n=3, are for (m(0), m(1), m(2), m(3), m(4)): (1,1,0,0,1), (1,0,1,1,0), (0,2,0,1,0) and (0,1,2,0,0).
		

References

• G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 240, 242-3.

Links

• Eric Weisstein's World of Mathematics, q-Binomial Coefficient.

Crossrefs

Cf. A000012 (as triangle for m=1), A008967 (m=2), A047971 (m=3).

Cf. A000332, A016813.

Programs

Mathematica

a[0, 0] = 1; a[n_, k_] := SeriesCoefficient[ QBinomial[n+4, 4, q], {q, 0, k}]; Table[a[n, k], {n, 0, 6}, {k, 0, 4*n}] // Flatten (* Jean-François Alcover, Dec 04 2013 *)

Formula

T(n,k) = [q^k] [x^n](1/Product_{j=0..4} (1-x*q^j)), n >= 0, 0 <= k <= 4*n.

T(n,k) = [q^k]([n+4,4]_q), n >= 0, 0 <= k <= 4*n.

See the comments above.