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A377418
a(n) is the smallest integer k whose set of divisors contains exactly n triples (x,y,z) of distinct divisors considered as integer-sided triangles with integer areas, or 0 if no such k exists.
0
60, 120, 240, 360, 960, 720, 3480, 1440, 1680, 2880, 6600, 2520, 4200, 10440, 5460, 6240, 4680, 5040, 20400, 7800, 18360, 17160, 26520, 10080, 47040, 9360, 15120, 10920, 55080, 20160, 15600, 16380, 34320, 33600, 18720, 27300, 165240, 53040, 37800, 25200, 21840
OFFSET
1,1
COMMENTS
We observe that this sequence is a subsequence of A169823: a(n) == 0 (mod 60).
The area A of a triangle whose sides have lengths x, y, and z is given by Heron's formula: A = sqrt(s*(s-x)*(s-y)*(s-z)), where s = (x+y+z)/2.
EXAMPLE
a(3) = 240 because there are 3 triples of divisors (3, 4, 5), (6, 8, 10) and (12, 16, 20) with integer areas 36, 576, 9216 respectively (Pythagorean triples). The first triple is primitive.
a(9)=1680 because there are 9 triples of divisors (3,4,5), (6,8,10), (7,15,20), (12,16,20), (14,30,40), (21,28,35), (28,60,80), (42,56,70), (84,112,140) with 5 Pythagorean triples : (3,4,5), (6,8,10), (21,28,35), (42,56,70), (84,112,70). The other 4 triangles are arbitrary.
MAPLE
with(numtheory):
for n from 1 to 41 do:
ii:=0:
for m from 4 to 10^7 while(ii=0) do:it:=0:
d:=divisors(m):n0:=nops(d):
for i from 2 to n0-2 do:
for j from i+1 to n0 do:
for k from j+1 to n0 do:
x:=d[i]:y:=d[j]:z:=d[k]:s:=(x+y+z)/2:A:=s*(s-x)*(s-y)*(s-z):
if A>0 and sqrt(A)=floor(sqrt(A)) then it:=it+1:else
fi:
od:
od:
od:
if it=n then printf(`%d %d \n`, it, m):ii:=1:
else fi:
od:
od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 27 2024
STATUS
approved