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%I #7 Jan 05 2014 10:54:52
%S 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,
%T 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,
%U 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
%N Number of totally symmetric 6-dimensional partitions of n.
%C 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.
%C 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.
%C 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.
%C In fact the first term of the sequence with n=2 mod 3, and which is nonzero is a(233) = 1
%H Graham H. Hawkes, <a href="/A234954/b234954.txt">Table of n, a(n) for n = 1...200</a>
%K nonn
%O 1,58
%A _Graham H. Hawkes_, Jan 01 2014
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