A219238 Coefficient table for the first differences of table A047971: Coefficients of the difference of Gauss polynomials [n+3,3]_q - [n+2,3]_q.

1, 0, 1, 1, 1, 0, 0, 1, 1, 2, 1, 1, 0, 0, 0, 1, 1, 2, 2, 2, 1, 1, 0, 0, 0, 0, 1, 1, 2, 2, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 4, 4, 3, 3, 2, 2, 1, 1

Offset: 0

Author

Wolfdieter Lang, Dec 06 2012

Keywords

• nonn
• easy
• tabf

Comments

The row lengths sequence is A016777 (3*n+1). The sum for row n is A000217(n+1) = binomial(n+2,2).

The coefficients of the Gauss polynomial [n+3,3]_q are given in A047971.

a(n,k) = [q^k]([n+3,3]_q - [n+2,3]_q). One can use the identity [n+3,3]_q - [n+2,3]_q = q^n*[n+2,2]_q (see the Andrews reference given in A047971, p. 35, (3.3.3)). Therefore, the present array is obtained from A008967 after a shift of row n by n units to the right, inserting zeros for the first n entries.

The o.g.f. of the row polynomials in q of degree 3*n is 1/((1-q)*(1-q^2)*(1-q^3)) (multiply the o.g.f. of A047971 by (1-z)). a(n,k) determines therefore the number of partitions of k with precisely n parts, each <= 3. Alternatively, a(n,k) determines the number of partitions of k with at most 3 parts, with each part <= n but not each part <= (n-1), i.e., part n, maybe more than once, is present besides possibly smaller ones.

Examples

			The table a(n,k) begins:
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18...
0:  1
1:  0  1  1  1
2:  0  0  1  1  2  1  1
3:  0  0  0  1  1  2  2  2  1  1
4:  0  0  0  0  1  1  2  2  3  2  2  1  1
5:  0  0  0  0  0  1  1  2  2  3  3  3  2  2  1  1
6:  0  0  0  0  0  0  1  1  2  2  3  3  4  3  3  2  2  1  1
...
Row n=1 is 0,1,1,1 because [3,2]_q = 1 + q + q^2 and the coefficient of q^{-1} is 0, the one of q^0 is 1, the one of q^1 is 1 and the one of q^2 is 1. A shift of row n=1 of A008967 by one unit to the right.
a(n,k) = 0 if n > k because a partition of k never has more than k parts.
a(n,k) = 0 if k > 3*n because there is no partition of 3*n+m, with m >= 1, and exactly n parts, each <= 3.
a(2,4) = 2 because the partitions of 4 with 2 parts are 1,3 and 2,2, and the parts in both are <= 3.
a(2,4) = 2 because the partitions of 4 with number of parts <= 3, each <= 2, are 2,2 and 1,1,2, and part 2 is present in both of them. Note the conjugacy of partitions 1,3 and 1,1,2.
		

Crossrefs

Cf. A047971, A008967 (with shifted rows).

Formula

a(n,k) = [q^k]([n+3,3]_q - [n+2,3]_q), = [q^(k-n)] [n+2,2]_q , n >= 0, 0 <= k <= 3*n. For the Gauss polynomial (q-binomial) [n+m,m]_q = [m+n,n]_q see a comment on A219237 where also the Andrews reference and a link to Mathworld is found.