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A229098
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Smallest area A of Heron triangles with sides (a, b, c) in arithmetic progression of the form b - d(n), b, b + d(n), where d(n) = A091998(n) = 12*n +- 1.
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1
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6, 156, 126, 546, 3750, 7350, 570, 1176, 14406, 2046, 3216, 4740, 1554, 3354, 43350, 54150, 6180, 3924, 17556, 84966, 3294, 24174, 106134, 7446, 126150, 144150, 28236, 33174, 21294, 10374, 6006, 9264, 16716, 247254, 252150, 277350, 282534, 55944, 75894, 26676
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OFFSET
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1,1
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COMMENTS
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According to the reference, d(n) is congruent (mod 12) to 1 or -1.
Let the sides be b - d, b, b + d where 1 <= d <= b. Then the semiperimeter s = 3b/2 and by Heron's formula, the area is A = b*sqrt(3*(b^2 - 4*d^2))/4.
The following table gives the first values (d(n), a, b, c, A):
+------+-----+-----+-----+-------+
| d(n) | a | b | c | A |
+------+-----+-----+-----+-------+
| 1 | 3 | 4 | 5 | 6 |
| 11 | 15 | 26 | 37 | 156 |
| 13 | 15 | 28 | 41 | 126 |
| 23 | 29 | 52 | 75 | 546 |
| 25 | 75 | 100 | 125 | 3750 |
| 35 | 105 | 140 | 175 | 7350 |
| 37 | 39 | 76 | 113 | 570 |
| 47 | 51 | 98 | 145 | 1176 |
| 49 | 147 | 196 | 245 | 14406 |
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LINKS
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EXAMPLE
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a(2) = 156 is in the sequence because d(2) = A091998(2) = 11 and (a, b, c) = (15, 26, 37) => the semiperimeter is (15 + 26 + 37)/2 = 39, and A = sqrt(39*(39-15)*(39-26)*(39-37)) = 156.
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MAPLE
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with(numtheory):u:=0:nn:=1000:lst:={1}:for k from 1 to 10 do:x:=12*k-1:y:=12*k+1:lst:=lst union {x} union {y}:od:for n from 1 to 20 do:ii:=0:d:=lst[n]:for b from 1 to nn while(ii=0)do:s:= b*sqrt(3*(b^2-4*d^2))/4:if s>0 and s=floor(s) then ii:=1:u:=u+1:printf ( "%d %d %d %d \n", u, d, b, s):else fi:od:od:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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