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Theorem (I. N. Ianakiev): There are infinitely many such numbers. Proof: Any A001110(2n+1), for n>0, is such a number as A001110(2n+1) = (2a+1)^2+(4a^2+4a)(4a^2+4a+1)(1/2), where a = (A002315(n)-1)(1/2). Note: other numbers, not of the form A001110(2n+1), e.g. A001110(6), are also in the sequence (see the example below).

Every term is divisible by its digital root (A010888). - Ivan N. Ianakiev, Oct 17 2013

Theorem (I. N. Ianakiev): There are infinitely many such numbers. Proof: There are infinitely many square triangular numbers (A001110) and every (2t+1)-th of them is odd because A001110(0)=0, A001110(1)=1 and A001110(n)=34*a(n-1)-a(n-2)+2, for n>=2. Any sqrt(A001110(2t+1)) is odd (i. e. is in A005408) and can be written as p^2-q^2 because A005408(n)=A000290(n+1)-A000290(n). The unique values of p and q (p>q>0) for each sqrt(A001110(2t+1)) generate (when t>0) a unique Pythagorean triple with a unique hypotenuse (a=p^2-q^2, b=2pq, c=p^2+q^2). Therefore, there are infinitely many such hypotenuses squared.