A382932 a(n) is the altitude of the Pythagorean triangle (A046083(A382931(n)), A046084(A382931(n)), A009000(A382931(n))).

12, 24, 36, 48, 60, 72, 60, 84, 96, 108, 120, 132, 120, 144, 156, 120, 168, 180, 192, 204, 216, 228, 240, 180, 252, 264, 276, 240, 288, 300, 168, 312, 324, 240, 336, 348, 360, 372, 384, 396, 420, 300, 408, 360, 420, 432, 444, 456, 468, 480, 360, 492, 504, 516

Offset: 1

Author

Felix Huber, Apr 13 2025

Keywords

• nonn

Comments

All terms are divisible by 12. Proof: (Start)

Let (a, b, c) be a primitive Pythagorean triple. Since gcd(a, b, c) = 1, all and only the Pythagorean triples (k*c*a, k*c*b, k*c^2) have an integer altitude h = (k*c*a*k*c*b)/(k*c^2) = k*a*b, where k is a positive integer.

With a = p^2 - q^2 and b = 2*p*q follows h = 2*k*p*q*(p^2 - q^2) = k*2*p*q*(p + q)*(p - q), where p > q > 0, gcd(p,q) = 1 and p or q is even.

It is to show that p*q*(p + q)*(p - q) is divisible by 6. Since p or q is divisible by 2, it remains to show that p*q*(p + q)*(p - q) is divisible by 3.

If 3 is a divisor of p or q, p*q is divisible by 3. If p mod 3 = 1 and q mod 3 = 2 or p mod 3 = 2 and q mod 3 = 1, then p + q is divisible by 3. If p mod 3 = q mod 3 = 1 or p mod 3 = q mod 3 = 2, then p - q is divisible by 3.

It follows that all terms are divisible by 12. (End)

Examples

			a(1) = 12 because the Pythagorean triangle (A046083(A382931(1)), A046084(A382931(1)), A009000(A382931(1))) = (A046083(7), A046084(7), A009000(7)) = (15, 20, 25) has the integer altitude 15*20/25 = 12.
		

Links

• Felix Huber, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Pythagorean Triple

Crossrefs

Cf. A008594, A009000, A046083, A046084, A382931.

Programs

Maple

A382932:=proc(H) # All hypotenuses <= H.
    local a,b,c,k,p,q,L,M;
    L:=[];
    M:=[];
    for p from 2 to floor(sqrt(H-1)) do
        for q to min(p-1,floor(sqrt(H-p^2))) do
            if gcd(p,q)=1 and is(p-q,odd) then
                a:=p^2-q^2;
                b:=2*p*q;
                c:=p^2+q^2;
                for k to iquo(H,c) do
                    L:=[op(L),[k*c,k*max(a,b),k*a*b/c]]
                od
            fi
        od
    od;
    L:=sort(L);
    for k to nops(L) do
        if is(L[k,3],integer) then
           M:=[op(M),L[k,3]]
        fi
    od;
    return op(M)
end proc;
A382932(1075);

Formula

a(n) = A046083(A382931(n))*A046084(A382931(n))/A009000(A382931(n)).