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A024162
Number of incongruent integer-sided triangles with sides a,b,c, a<b<c, a+b+c=n such that gcd(a,b,c)=1.
1
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 1, 3, 2, 4, 2, 5, 4, 7, 4, 8, 6, 10, 6, 11, 9, 14, 9, 16, 12, 18, 12, 21, 15, 24, 16, 25, 20, 30, 19, 33, 25, 33, 25, 40, 28, 44, 30, 44, 36, 52, 34, 55, 42, 56, 42, 65, 44, 70, 49, 67, 56, 78, 52, 85, 64, 83, 63, 96, 66
OFFSET
0,14
REFERENCES
A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 8, #30 (First published: San Francisco: Holden-Day, Inc., 1964)
FORMULA
For n=0..11 (mod 12), a(n) is respectively (n^2 + 6n - 7)/48, (n^2 - 4)/48, (n^2 + 6n + 21)/48, (n^2 - 16)/48, (n^2 + 6n - 7)/48, (n^2 + 12)/48, (n^2 + 6n + 5)/48, (n^2 - 16)/48, (n^2 + 6n + 9)/48, (n^2 - 4)/48, (n^2 + 6n + 5)/48
From Felix Huber, Sep 02 2025: (Start)
A024161(n) <= a(n) <= A051493(n).
Conjecture: a(2*k-1) >= a(2*k). (End)
EXAMPLE
The a(18) = 2 triangles are (3, 7, 8) and (5, 6, 7). - Felix Huber, Sep 02 2025
MAPLE
A024162:=proc(n)
local a, x, z;
a:=0;
for x to n/3-1 do
for z from floor((n-x)/2)+1 to min((n-1)/2, n-2*x-1) do
if igcd(x, n-x-z, z)=1 then
a:=a+1
fi
od
od;
return a
end proc;
seq(A024162(n), n=0..72); # Felix Huber, Sep 02 2025
CROSSREFS
KEYWORD
nonn,nice,easy
EXTENSIONS
Reference and formulae from Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 27 2000
Sequence revised, name clarified and a(0) = 0 prepended by Felix Huber, Sep 02 2025
STATUS
approved