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%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
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