*** caution ***

Now, this Anniversary Coin has been stolen by someone.

***

Can you figure out whether the Anniversary coin is heavier or lighter using a balance scale at most six times?

]]>One way to produce algebraic integers P whose powers are quite close to integers is to observe that the sums of the powers of the conjugates of P are all integral. In particular, if all the other conjugates of P have absolute value less than one, then the closest integer to P^n (for large n) satisfies a linear recurrence relation. For example, if you take P = (sqrt(5)+1)/2 to be the golden ratio, and Q = (-sqrt(5)+1)/2 to be the conjugate of P, then

PQ = -1, |P| = 1.61803… |Q| = 0.61803…

And now the sequence P^n + Q^n consists entirely of integers: 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, and so on. These are the Lucas numbers L_n. Now let’s see what happens if we square both sides. We get

P^(2n) + 2 P^n Q^n + Q^(2n) = 1,9,16,49,… = (L_n)^2.

But as observed, PQ = -1 so P^n Q^n = (-1)^n, and Q^(2n) is a (positive!) real number less than one. Hence

P^(2n) + 2 (-1)^n + real number less than one = (L_n)^2.

But that means the ceiling of P^(2n) differs from (L_n)^2 by 2 or -2, and so you can take A = P^2 in the problem. (If you want the sign of the “2” to be the same, you can take A = P^4 instead.)

This argument works equally well if P is any unit in a real quadratic number field, and A=P^2. So you could also take A = (1 + sqrt(2))^2 = (3+2 sqrt(2)) or (2 + sqrt(3))^2 = (7 + 2 sqrt(3)), etc.

]]>I was looking at OEIS sequence A131481 – numbers of treelike polyiamonds, aka numbers of polyiamonds that have maximal perimeter, which you posted in 2007.

In my opinion, the data is correct but shifted by 1.

Maybe the problem is due to the offset = 0, but in any case, to take as an example a(6), it should be 11, whereas you give a(5) = 11.

Did you really want offset = 0?

Or have I read everything completely wrong?

john ]]>