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A062890
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Number of quadrilaterals that can be formed with perimeter n. In other words, number of partitions of n into four parts such that the sum of any three is more than the fourth.
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14
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0, 0, 0, 0, 1, 1, 1, 2, 3, 4, 5, 7, 8, 11, 12, 16, 18, 23, 24, 31, 33, 41, 43, 53, 55, 67, 69, 83, 86, 102, 104, 123, 126, 147, 150, 174, 177, 204, 207, 237, 241, 274, 277, 314, 318, 358, 362, 406, 410, 458, 462, 514, 519, 575, 579, 640, 645, 710
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OFFSET
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0,8
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COMMENTS
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Partition sets of n into four parts (sides) such that the sum of any three is more than the fourth do not uniquely define a quadrilateral, even if it is further constrained to be cyclic. This is because the order of adjacent sides is important. E.g. the partition set [1,1,2,2] for a perimeter n=6 can be reordered to generate two non-congruent cyclic quadrilaterals, [1,2,1,2] and [1,1,2,2], where the first is a rectangle and the second a kite. - Frank M Jackson, Jun 29 2012
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,0,0,-1,1,-1,1,1,-1).
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FORMULA
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G.f.: x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-8) + a(n-9) - a(n-10) + a(n-11) + a(n-12) - a(n-13).
a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} sign(floor((i+j+k)/(n-i-j-k+1))). (End)
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EXAMPLE
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a(7) = 2 as the two partitions are (1,2,2,2), (1,1,2,3) and in each sum of any three is more than the fourth.
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MATHEMATICA
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CoefficientList[Series[x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)), {x, 0, 60}], x] (* Frank M Jackson, Jun 09 2017 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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