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%I #24 Apr 30 2016 12:19:45
%S 1,1,1,2,2,2,1,3,2,2,2,3,2,4,2,3,2,4,1,5,2,2,3,4,3,4,2,4,3,5,1,6,3,2,
%T 3,5,1,6,3,5,3,4,2,5,3,3,3,6,2,8,2,3,4,5,3,7,2,3,3,8,2,5,4,4,4,4,2,8,
%U 2,6,2,7,2,7,3,3,5,5,1,8,4,3,5,6,4,7,2,6,2,8,3,7,3,3,5,7,1,8,3,7
%N Number of haystack arrangements with n bales.
%C A popular way of arranging cylindrical bales of hay is for the first row to be a rectangle with dimensions (x,y), and the next row to be (x-1, y-1), and so on. The height is h where h<=min(x,y). This sequence gives the number of distinct (y>=x) arrangements which result in n bales.
%F The size of the haystack s(x,y,h) = h(6xy-3(x+y)(h-1) + (h-1)(2h-1))/6 (courtesy of Sascha Troscheit).
%e a(14)=4 since there are 4 haystacks with 14 bales: (x=3,x=3,h=3), (x=2,x=5,h=2), (x=2,y=7,h=1), (x=1,y=14,h=1).
%o (PARI) s(x,y,h) = h*(6*x*y-3*(x+y)*(h-1) + (h-1)*(2*h-1))/6 ;
%o a(n) = my(nbc); for (y=1, n, for(x=1, y, for (h=0, min (x,y), if (s(x,y,h) == n, nbc++)))); nbc; \\ _Michel Marcus_, Apr 30 2016
%K nonn
%O 1,4
%A _Paul Richards_, Aug 29 2015
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