+5: 2+3 -> 3 ; 4+1 -> 3 ; ---> 3

+6: 3+3 -> 3 ; 4+2 -> 3 ; 1+5 -> 4

+7: 1,2,3,4,7 -> 4 ; 1,2,3,6,7 -> 4

+9: 1,2,4,8,9 -> 4 ; 1,2,3,6,9 -> 4 ; 1,2,4,5,9 -> 4

+11: 1,2,4,8,10,11 -> 5 ; 1,2,3,6,9,11 -> 5 ; 1,2,4,5,10,11 -> 5 ;

+12: 1,2,3,6,12 -> 4 ; 1,2,4,8,12 -> 4 [3 * 2 * 2]

+13: 1,2,4,8,12,13 -> 5 ; 1,2,3,6,12,13 -> 5

+14: 7*2 -> 5 ; 1,2,4,8,12,14 -> 5

+15: 1,2,3,6,9,15 -> 5 ; 1,2,3,5,10,15 -> 5

+rank(2*x+1) >= rank(2*x).

+Proof: let's assume that x is the minimal natural number for which

+rank(2*x+1) < rank(2*x). If its best composition was "(2*x)+(1)" then

+2*x would have a lower rank. Therefore it is i+j. Now since 2*x+1 is odd,

+it means one of i,j is even and the other odd. Without loss of generality,

+let's assume that i is even and j is odd.

+Let's look at the composition C[2*x+1] at one point we add 1 to an even number

+Let's not add it there. If we can still form 2*x with the lower rank then we

+have a contradiction. If, OTOH, we do something with [2*a'+1], like add it to

+another number then either:

+1. Instead of doing [2*a'+1]+[2*b'+1] we can do [2*a']+[2*b']+2 to get the

+same sum with the same, or lesser, effort.

+2. If we multiply [2*a'+1] by two to get [2*a'+1]*2, we can do

+[2*a']*2+2 and get it at the same effort.

+Therefore, the rank of 2*x+1 must be greater or equal to that of 2*x.

+For every x \in N : rank(2*x) == rank(x)+1 and its best composition is that of

+Proof: let's call r = rank(x). So in order to beat the "x+x" composition,

+we need to find two integers, i and j <> x, so that i+j=x+x and

+rank(i)+rank(j)-rank(i /\ j) <= r-1 . Now, if i and j are both even,

+the we can take the "i/2+j/2" composition to find a better composition for

+x, and so yield a lower r (which is reductio ad absurdum). Therefore, i

+and j are both odd (because an odd number plus an even number cannot be

+Let's assume that i < j (if i=j then i=j=x). If i = 1 then according to

+Lemma 1 then rank(j-1) <= rank(j) and we can use "(j-1)+(2)" instead

+to yield a minimal sum which is a contradiction.

+Let's look at "(i-1)+(j-1)"

+Now since rank(i-1) <= rank(i) and rank(j-1) <= rank(j)

+Now, in order to form an odd number one has to add 1 to an even power

+of 2 such as 2, 4, 8, 16, etc (which clearly are minimally composed).

+If C[2*x+1] (where C is the composition) is {1,2,.....a2,a1, 2*x+1}

+then a2 != a1 or else 2*x+1 will be odd. So a1 > x. Now, let's take the

+identical set only with a1-1 - {1,2,.....a2,a1-1,2*x} . We need to compose

+a1-1 somehow and make sure that (2*x+1-a1) is still a composing number.

+a1 = a[n] + a[m] -> a1-1 = a[n] + a[m]-1.

+Let's take "(i)+(j-1)". Since j-1 < 2*x, then rank(j-1) <= rank(j). So

+rank(i) + rank(j-1) <= rank(i) + rank(j) . But it is possible i and j

+have some common components. Let's mark these components as {a1,a2,a3,a4}

+It is clear that a1 is 1 and a2 = 2, because these start all compositions.

+Since j-1 < j then the number of elements in the j-1 set is lesser or