Ulam number

Examples

As a consequence of the definition, 3 is an Ulam number (1 + 2); and 4 is an Ulam number (1 + 3). (Here 2 + 2 is not a second representation of 4, because the previous terms must be distinct.) The integer 5 is not an Ulam number, because 5 = 1 + 4 = 2 + 3. The first few terms are :1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, 38, 47, 48, 53, 57, 62, 69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126, 131, 138, 145, 148, 155, 175, 177, 180, 182, 189, 197, 206, 209, 219, 221, 236, 238, 241, 243, 253, 258, 260, 273, 282, ... .

There are infinitely many Ulam numbers. For, after the first n numbers in the sequence have already been determined, it is always possible to extend the sequence by one more element: U**n−1 + U**n is represented as a sum of two of the first n numbers, so either it is a term of the sequence or it can be expressed in a different way as a sum of two elements. But since all other sums of two elements among the first n are less than U**n−1 + U**n, there must be an element of the sequence between U**n and U**n−1 + U**n. The next element is therefore the smallest of the uniquely representable numbers that exceed U**n.

Ulam is said to have conjectured that the numbers have zero density, but they seem to have a density of approximately 0.07398.

Properties

Apart from 1 + 2 = 3 any subsequent Ulam number cannot be the sum of its two prior consecutive Ulam numbers. :Proof: Assume that for n 2, U**n−1 + U**n = U**n+1 is the required sum in only one way; then so does U**n−2 + U**n produce a sum in only one way, and it falls between U**n and U**n+1. This contradicts the condition that U**n+1 is the next smallest Ulam number.

For n 2, any three consecutive Ulam numbers (U**n−1, U**n, U**n+1) as integer sides will form a triangle.

:Proof: The previous property states that for n 2, U**n−2 + U**n ≥ U**n + 1. Consequently U**n−1 + U**n U**n+1 and because U**n−1 n n+1 the triangle inequality is satisfied.

The sequence of Ulam numbers forms a complete sequence. :Proof: By definition U**n = U**j + U**k where j n with n 3, the greatest value that U**j can have is U**n−3 and the greatest value that U**k can have is U**n−1. :Hence U**n ≤ U**n−1 + U**n−3 n−1 and U**1 = 1, U**2 = 2, U**3 = 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 U**j such that n ≤ U**j :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 U**j−1 ≤ n ≤ U**j. From the proof above for n 3, U**j ≤ U**j−1 + U**j−3 j−1. Therefore n ≤ U**j j−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 = U**j an Ulam number then it is possible that i + 1 is the next Ulam number U**j+1. Now consider i + 2, this cannot be the next Ulam number U**j+2 because it is not a unique sum of two previous terms. i + 2 = U**j+1 + U1 = U**j + 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 U**n+1 ≤ U**n + U**n−2 and in any five consecutive positive integers at most two can be Ulam numbers, it can be stated that :n−7 ≤ U**n ≤ N**n+1 for n 0, where N**n are the numbers in Narayana’s cows sequence: 1,1,1,2,3,4,6,9,13,19,... with the recurrence relation N**n = N**n−1 +N**n−3 that starts at N0.

Hidden structure

It has been observed that the first 10 million Ulam numbers satisfy \cos{(2.5714474995 , U_n)} except for the four elements \left{2,3,47,69\right} (this has now been verified for the first 10^9 Ulam numbers). Inequalities of this type are usually true for sequences exhibiting some form of periodicity but the Ulam sequence does not seem to be periodic and the phenomenon is not understood. It can be exploited to do a fast computation of the Ulam sequence (see External links).

Generalizations

The idea can be generalized as (u, v)-Ulam numbers by selecting different starting values (u, v). A sequence of (u, v)-Ulam numbers is regular if the sequence of differences between consecutive numbers in the sequence is eventually periodic. When v is an odd number greater than three, the (2, v)-Ulam numbers are regular. When v is congruent to 1 (mod 4) and at least five, the (4, v)-Ulam numbers are again regular. However, the Ulam numbers themselves do not appear to be regular.

A sequence of numbers is said to be s-additive if each number in the sequence, after the initial 2s terms of the sequence, has exactly s representations as a sum of two previous numbers. Thus, the Ulam numbers and the (u, v)-Ulam numbers are 1-additive sequences.

If a sequence is formed by appending the largest number with a unique representation as a sum of two earlier numbers, instead of appending the smallest uniquely representable number, then the resulting sequence is the sequence of Fibonacci numbers.

Notes

References

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References

1. {{harvs. Ulam. (1964a). txt
2. {{harvtxt. Recaman. 1973 gives a similar argument, phrased as a [[proof by contradiction]]. He states that, if there were finitely many Ulam numbers, then the sum of the last two would also be an Ulam number – a contradiction. However, although the sum of the last two numbers would in this case have a unique representation as a sum of two Ulam numbers, it would not necessarily be the smallest number with a unique representation.
3. The statement that Ulam made this conjecture is in OEIS {{OEIS2C. Ulam. 1964a, and in {{harvtxt. Ulam. 1964b he poses the question of determining its density without conjecturing a value for it. {{harvtxt. Recaman. 1973 repeats the question from {{harvtxt. Ulam. 1964b of the density of this sequence, again without conjecturing a value for it.
4. OEIS {{OEIS2C
5. {{harvtxt. Recaman. 1973
6. OEIS {{OEIS2C
7. Philip Gibbs and Judson McCranie. (2017). "The Ulam Numbers up to One Trillion".
8. {{harvtxt. Steinerberger. 2015
9. {{harvtxt. Queneau. Finch. 1992 conjectured the extension of this result to all odd ''v'' greater than three, and this conjecture was proven by {{harvtxt. Schmerl. Spiegel. 1994. The regularity of the (4, ''v'')-Ulam numbers was proven by {{harvtxt. Cassaigne. Finch. 1995.
10. {{harvtxt. Queneau. 1972.
11. {{harvtxt. Finch. 1992.