A234954 Number of totally symmetric 6-dimensional partitions of n.

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

Offset: 1

Author

Graham H. Hawkes, Jan 01 2014

Keywords

• nonn

Comments

We can think of the points of a totally symmetric partition of n, say p, as occurring in classes, where two points are in the same class iff one point is a given by a permutation of the coordinates of the other. Call the number of distinct points in a class the size of that class.

The only classes of points in a 6-dimensional totally symmetric partition, p, of n, which do not have class size divisible by 3 are composed of points of the form (x,x,x,x,x,x) or (x,x,x,y,y,y) (or any permutation of these coordinates). The former has class size 1, the latter, class size 20.

For n=2 mod 3, a(n)=0 for the first 232 terms. Indeed, suppose n<233 and n=2 mod 3 and p partitions n in 6 dimensions. If j is the number of points of the form (x,x,x,x,x,x) in p, and k is the number of points of the form (x,x,x,y,y,y) in p, then we must have j+2k = 2 mod 3. Now j>0 because (1,1,1,1,1,1) must be a point of p. If j=1, we have k=2 mod 3, so that k>=2. In this case, the minimum size of n occurs when k=2 and the two points of the form (x,x,x,y,y,y) are (2,2,2,1,1,1) and (3,3,3,1,1,1). In this case, n=233. If j=2, we have k=0 mod 3. But since j=2,(2,2,2,2,2,2) is a point of p. Thus, so is(2,2,2,1,1,1). Hence, k>0, whence k>=4. In particular, k>=2 so that n>233. If j>=3, then (3,3,3,3,3,3) is a point of p, in which case n>729=3^6.

In fact the first term of the sequence with n=2 mod 3, and which is nonzero is a(233) = 1

Links

• Graham H. Hawkes, Table of n, a(n) for n = 1...200