OFFSET
1,2
LINKS
Vladimir Letsko, Mathematical Marathon, problem 192 (in Russian).
FORMULA
a(n) = Sum_{j=0..floor(n*(1 - sqrt(2)/2))} (n - j - floor(sqrt(2*j*n - j^2))). - Anton Nikonov, Oct 06 2014
a(n) = (1/8)*(-4*ceiling((n - 1)/sqrt(2)) + 4*n^2 - A000328(n) + 1), n > 1. - Mats Granvik, May 23 2015
EXAMPLE
a(3) = 3 because there are 3 integer-sided acute triangles with largest side 3: (1,3,3); (2,3,3); (3,3,3).
MAPLE
tr_a:=proc(n) local a, b, t, d; t:=0:
for a to n do
for b from max(a, n+1-a) to n do
d:=a^2+b^2-n^2:
if d>0 then t:=t+1 fi
od od;
t; end;
MATHEMATICA
a[ n_] := Length @ FindInstance[ n >= b >= a >= 1 && n < b + a && n^2 < b^2 + a^2, {a, b}, Integers, 10^9]; (* Michael Somos, May 24 2015 *)
PROG
(PARI) a(n) = sum(j=0, n*(1 - sqrt(2)/2), n - j - floor(sqrt(2*j*n - j^2))); \\ Michel Marcus, Oct 07 2014
(PARI) {a(n) = sum(j=0, n - sqrtint(n*n\2) - 1, n - j - sqrtint(2*j*n - j*j))}; /* Michael Somos, May 24 2015 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Letsko, Sep 20 2014
STATUS
approved