OFFSET
1,1
COMMENTS
a(1) = A011945(1).
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 |
LINKS
J. A. MacDougall, Heron Triangles With Sides in Arithmetic Progression, School of Mathematical and Physical Sciences, University of Newcastle, NSW, Australia 2308, February 2, 2005.
Kival Ngaokrajang, Illustration of initial terms
EXAMPLE
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.
MAPLE
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:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Sep 13 2013
STATUS
approved