+ Un > Un+1 and because Un−1 < Un < Un+1 the triangle inequality is satisfied. The sequence of Ulam numbers forms a complete sequence. Proof: By definition Un = Uj + Uk where j < k < n and is the smallest integer that is the sum of two distinct smaller Ulam numbers in exactly one way. This means that for all Un with n > 3, the greatest value that Uj can have is Un−3 and the greatest value that Uk can have is Un−1. Hence Un ≤ Un−1 + Un−3 < 2Un−1 and U1 = 1, U2 = 2, U3 = 3. This is a sufficient condition for Ulam numbers to be a complete sequence. For every integer n > 1 there is always at least one Ulam number Uj such that n ≤ Uj < 2n. Proof: It has been proved that there are infinitely many Ulam numbers and they start at 1. Therefore for every integer n > 1 it is possible to find j such that Uj−1 ≤ n ≤ Uj. From the proof above for n > 3, Uj ≤ Uj−1 + Uj−3 < 2Uj−1. Therefore n ≤ Uj < 2Uj−1 ≤ 2n. Also for n = 2 and 3 the property is true by calculation. In any sequence of 5 consecutive positive integers {i, i + 1,..., i + 4}, i > 4 there can be a maximum of 2 Ulam numbers. Proof: Assume that the sequence {i, i + 1,..., i + 4} has its first value i = Uj an Ulam number then it is possible that i + 1 is the next Ulam number Uj+1. Now consider i + 2, this cannot be the next Ulam number Uj+2 because it is not a unique sum of two previous terms. i + 2 = Uj+1 + U1 = Uj + U2. A similar argument exists for i + 3 and i + 4. Inequalities Ulam numbers are pseudo-random and too irregular to have tight bounds. Nevertheless from the properties above, namely, at worst the next Ulam number Un+1 ≤ Un + Un−2 and in any five consecutive positive integers at most two can be Ulam numbers, it can be stated that ≤ Un ≤ Nn+1 for n > 0, where Nn are the numbers in Narayana’s cows sequence: 1,1,1,2,3,4,6,9,13,19,... with the recurrence relation Nn = Nn−1 +Nn−3 that starts at