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A260372
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Number of haystack arrangements with n bales.
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0
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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, 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, 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
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OFFSET
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1,4
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COMMENTS
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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.
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LINKS
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Table of n, a(n) for n=1..100.
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FORMULA
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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).
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EXAMPLE
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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).
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PROG
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(PARI) s(x, y, h) = h*(6*x*y-3*(x+y)*(h-1) + (h-1)*(2*h-1))/6 ;
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
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CROSSREFS
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Sequence in context: A329028 A257079 A327567 * A037180 A241675 A349273
Adjacent sequences: A260369 A260370 A260371 * A260373 A260374 A260375
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KEYWORD
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nonn
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AUTHOR
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Paul Richards, Aug 29 2015
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STATUS
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approved
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