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A260372 Number of haystack arrangements with n bales. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..100.

FORMULA

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).

EXAMPLE

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).

PROG

(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

CROSSREFS

Sequence in context: A194290 A329028 A257079 * A037180 A241675 A091222

Adjacent sequences:  A260369 A260370 A260371 * A260373 A260374 A260375

KEYWORD

nonn

AUTHOR

Paul Richards, Aug 29 2015

STATUS

approved

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Last modified December 14 07:03 EST 2019. Contains 329978 sequences. (Running on oeis4.)