OFFSET
1,3
COMMENTS
Number of triples a,b,c with a <= b <= c < a+b, gcd(a,b,c) = 1 and c = n.
Dropping the requirement for side lengths to be relatively prime this sequence becomes A002620 (with a different offset). See the Sep 2006 comment in A002620. - Peter Munn, Jul 26 2017
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000
Stackexchange, Number of triples for which gcd(a,b,c)=1 and c=n, Feb 25 2014
FORMULA
Moebius transform of b(n) = floor((n+1)^2/4).
G.f.: (G(x)+x-x^2)/2, where G(x) = Sum_{k >= 1} mobius(k)*x^k*(1+2*x^k-x^(2*k))/(1-x^k)^2/(1-x^(2*k)).
MAPLE
with(numtheory):
a:= n-> add(mobius(n/d)*floor((d+1)^2/4), d=divisors(n)):
seq(a(n), n=1..60); # Alois P. Heinz, Oct 23 2013
MATHEMATICA
a[n_] := DivisorSum[n, Floor[(#+1)^2/4]*MoebiusMu[n/#]&]; Array[a, 60] (* Jean-François Alcover, Dec 07 2015 *)
PROG
(PARI) A123323(n)=sumdiv(n, d, floor((d+1)^2/4)*moebius(n/d)).
CROSSREFS
KEYWORD
AUTHOR
Franklin T. Adams-Watters, Sep 25 2006
STATUS
approved