A390820 - OEIS

A390820

Numbers k such that there is a primitive right triangle with perimeter + inradius = k^2.

19, 26, 43, 52, 64, 89, 94, 101, 104, 118, 151, 157, 166, 188, 191, 229, 236, 247, 254, 257, 274, 281, 331, 332, 344, 353, 358, 376, 433, 446, 457, 478, 494, 508, 509, 559, 569, 577, 586, 599, 608, 613, 628, 647, 689, 701, 712, 722, 727, 746, 757, 764, 767, 778, 808, 859, 871, 886, 892, 893, 922

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OFFSET

1,1

COMMENTS

Solutions to 2*m^2 + 3*m*n - n^2 = k^2 where m and n are coprime and one of m and n is even.

The first term that arises in more than one way is 2678, which comes from m=1681, n=322 and m=1769, n=178.

LINKS

Robert Israel, Table of n, a(n) for n = 1..4600

EXAMPLE

a(3) = 43 is a term because with m = 26 and n = 7, the primitive right triangle with sides a = m^2 - n^2 = 627, b = 2*m*n = 364 and c = m^2 + n^2 = 725 has perimeter a+b+c = 1716, inradius (a+b-c)/2 = 133 and perimeter + inradius = 1716 + 133 = 1849 = 43^2.

MAPLE

N:= 1000: # for terms <= N

R:= {}:

for m from 1 while m^2 < N^2 do

for n from 1+(m mod 2) to m-1 by 2 do

v:= 2*m^2 + 3*m*n - n^2; if v > N^2 then break fi;

if igcd(m, n) = 1 and issqr(v) then

y:= sqrt(v);

R:= R union {y}

od od:

sort(convert(R, list));

CROSSREFS