A247588 - OEIS

%I #27 Sep 05 2021 18:23:01

%S 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,

%T 90,96,104,111,117,126,132,140,147,154,165,172,183,189,198,210,219,

%U 229,237,247,260,270,282,292,302

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

%H Vladimir Letsko, <a href="http://dxdy.ru/post909787.html#p909787">Mathematical Marathon, problem 192</a> (in Russian).

%F 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

%F a(n) = (1/8)*(-4*ceiling((n - 1)/sqrt(2)) + 4*n^2 - A000328(n) + 1), n > 1. - _Mats Granvik_, May 23 2015

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

%p tr_a:=proc(n) local a,b,t,d;t:=0:

%p for a to n do

%p for b from max(a,n+1-a) to n do

%p d:=a^2+b^2-n^2:

%p if d>0 then t:=t+1 fi

%p od od;

%p t; end;

%t 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 *)

%o (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

%o (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 */

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

%K nonn

%O 1,2

%A _Vladimir Letsko_, Sep 20 2014