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A261491
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a(n) = ceiling(2 + sqrt(8*n-4)).
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5
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4, 6, 7, 8, 8, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 29
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) = minimal number of stones needed to surround area n in the middle of a Go board (infinite if needed).
The formula was constructed this way: when the area is in a diamond shape with x^2+(x-1)^2 places, it can be surrounded by 4x stones. So, a(1)=4, a(5)=8, a(13)=12 etc.
The positive solution to the quadratic equation 2x^2 - 2x + 1 = n is x = (2 + sqrt(8n-4))/4. And since a(n)=4x, the formula a(n) = 2 + sqrt(8n-4) holds for the positions mentioned. But incredibly also the intermediate results seem to match when the ceiling function is used.
The opposite of this would be an area of 1 X n; it demands the maximal number of stones, a(n) = 2 + 2n.
Equivalently, a(n) is the minimum (cell) perimeter of any polyomino of n cells. - Sean A. Irvine, Oct 17 2020
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LINKS
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FORMULA
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a(n) = ceiling(2 + sqrt(8*n-4)).
For n > 2, a(n) - a(n-1) = 1 if n is of the form 2*(k^2+k+1), 2*k^2 + 1 or (k^2+k)/2 + 1, otherwise 0. - Jianing Song, Aug 10 2021
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EXAMPLE
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Start with the 5-cell area that is occupied by 0's and surrounded by stones 1..8. Add those surrounding stones to the area, one by one. At points 1, 2, 4 and 6, the number of surrounding stones is increased; elsewhere, it is not.
Next, do the same with stones A..L. At points A, C, F and I, the number of surrounding stones is increased; elsewhere, it is not.
___D___
__A5C__
_B104E_
G30007J
_F206I_
__H8K__
___L___
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MATHEMATICA
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PROG
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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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STATUS
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approved
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