OFFSET
0,9
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.
T(n,k) is the number of subset sums of 4-element combinations of [1,2,..,n+4] equal to k+10. - David Cleaver, Jan 19 2026
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.
Wikipedia, Gaussian binomial coefficient.
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.
EXAMPLE
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).
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 *)
PROG
(PARI) row(n)={my(z=vector(0)); forsubset([n+4, 4], s, z=concat(z, vecsum(Vec(s)))); matreduce(z~)~[2, ]} \\ David Cleaver, Jan 19 2026
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Wolfdieter Lang, Dec 04 2012
STATUS
approved