A247588 Number of integer-sided acute triangles with largest side n.

1, 2, 3, 5, 6, 8, 11, 13, 15, 17, 21, 25, 27, 31, 34, 39, 43, 48, 52, 56, 63, 67, 73, 80, 84, 90, 96, 104, 111, 117, 126, 132, 140, 147, 154, 165, 172, 183, 189, 198, 210, 219, 229, 237, 247, 260, 270, 282, 292, 302

Offset: 1

Author

Vladimir Letsko, Sep 20 2014

Keywords

• nonn

Examples

			a(3) = 3 because there are 3 integer-sided acute triangles with largest side 3: (1,3,3); (2,3,3); (3,3,3).
		

Links

• Vladimir Letsko, Mathematical Marathon, problem 192 (in Russian).

Crossrefs

Cf. A046080, A002623, A224921, A247586, A247587, A247589.

Programs

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 *)

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 */

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