See http://www.research.att.com/~njas/sequences/A005318 and references therein ]]>

Call a set of +ve integers *distinctive* if every subset sums to a distinct value.

Call the maximum of a finite set its *max*.

Let m(k) be the minimum max over all distinctive k-sets (i.e. the sets of order k).

We are interested to track the distinctive k-sets which exhibit a max value = m(k).

It’s easy to see that:

m(1) = 1, exhibited by {1}

m(2) = 2, exhibited by {1,2}

m(3) = 4, exhibited by {1,2,4} or {2,3,4}

and slightly harder to see that:

m(4) = 7, exhibited by {3,5,6,7}

But what happens after that?

One clear pattern is emerging of powers of 2: {1}, {1,2}, {1,2,4}…

But for k=4, this pattern would give {1,2,4,8} and we can see that this is not optimal.

But how about extending the pattern which goes {1}, {1,2}, {2,3,4}…?

Suppose that we have a distinctive set where for all relevant i, the sum of any i members is always less than the sum of any i+1 members. We will call such a set *orderly*. It’s

pretty clear that {1}, {1,2}, {2,3,4} are all orderly.

What is less obvious is that from any orderly k-set, S, we can construct an infinite number of orderly (k+1)-sets.

PROOF BY CONSTRUCTION:

Write the 2^k subset-sums of S in numerical order, dividing them up into *segments* according to the order of the subset:

a_(0,1),

a_(1,1), a_(1,2), … a_(1,k),

a_(2,1), a_(2,2), … a_(2,k(k-1)/2),

…

a_(i,1), a_(i,2), … a_(i,C(k,i)),

…

a_(k,1)

If we made T=S U {0}, then T is not distinctive. Every subset-sum of T can be made in 2 ways, with and without 0.

But if we define U = T+n i.e.g U = {x | x-n in T} for some suitably large n yet to be defined, then there are two families of subset-sums of U:

a_(0,1),

n+a_(1,1), n+a_(1,2), … n+a_(1,k),

2n+a_(2,1), 2n+a_(2,2), … 2n+a_(2,k(k-1)/2),

…

in+a_(i,1), in+a_(i,2), … in+a_(i,C(k,i)),

…

kn+a_(k,1)

and:

n+a_(0,1),

2n+a_(1,1), 2n+a_(1,2), … 2n+a_(1,k),

3n+a_(2,1), 3n+a_(2,2), … 3n+a_(2,k(k-1)/2),

…

(i+1)n+a_(i,1), (i+1)n+a_(i,2), … (i+1)n+a_(i,C(k,i)),

…

(k+1)n+a_(k,1)

Now no two values from the same family are the same, because all that has happened in each family is that the segments have been spread further apart.

If n is suitably large, then there will be no overlaps across the two families. The segments from the two families will alternate in size.

Firstly:

in+a_(i-1,1), in+a_(i-1,2), … in+a_(i-1,C(k,i-1)) < in+a_(i,1), in+a_(i,2), … in+a_(i,C(k,i)),

as long as:

a_(i-1,C(k,i-1)) < a_(i,1)

which is always true, by the assumption that the original set was orderly.

But secondly:

in+a_(i,1), in+a_(i,2), … in+a_(i,C(k,i)), a_(i,C(k,i))-a_(i,1).

So if we set n*(k) = 1 + max{i | a_(i,C(k,i))-a_(i,1)}, then for any n >= n*(k), U is orderly. The old segments concatenate in pairs to make new ones:

in+a_(i-1,1), in+a_(i-1,2), … in+a_(i-1,C(k,i-1)) concatenated with in+a_(i,1), in+a_(i,2), … in+a_(i,C(k,i))

becomes:

b(i,1), b(i,2), … b(i,C(k+1,i))

where:

b(i,j) = in+a_(i-1,j), for j =< C(k,i-1)

b(i,j) = in+a_(i,j-C(k,i-1)), otherwise

END OF PROOF

Now we will always take n = n*(k), because we are looking for the minimal solution.

Let’s see this at work, then. m(3)=4, exhibited by (2,3,4). The subset sums are:

0,

2,3,4,

5,6,7,

9

n*(3) = 3 (the span of the largest segment, so the two families for subset-sums for the next level up:

0,

5,6,7,

11,12,13,

18

3,

8,9,10,

14,15,16,

21

These concatenate in pairs as:

0,

3,5,6,7,

8,9,10,11,12,13,

14,15,16,18,

21

We can read off the set exhibiting M(4)=7 from the second row.

Now let’s explore unknown terrain. n*(4)=6, so we make two families:

0,

9,11,12,13,

20,21,22,23,24,25,

32,33,34,36,

45

6,

15,17,18,19,

26,27,28,29,30,31,

38,39,40,42,

51

These concatenate in pairs as:

0,

6,9,11,12,13,

15,17,18,19,20,21,22,23,24,25,

26,27,28,29,30,31,32,33,34,36,

38,39,40,42,45,

51

Reading the second row, the set exhibiting M(5)=13 is {6,9,11,12,13}

Now for the next stage note that n*(5)=11 not 10. (There are gaps in the two longest segments at 16 & 35.)

We can then see that M(6)=24, exhibited by {11,17,20,22,23,24}.

24 is =< 26, so we have a constructive proof of the first part of the puzzle.

]]>Zeckendorf representation uses non-consecutive Fibonacci numbers. There are many way to represent a number as a sum of Fibonacci numbers: 9 = 8 + 1 = 5 + 3 + 1.

]]>Instead of:

a(n) ≥ log2n

one gets

a(n) ≥ loggn, g – golden ratio ]]>