Relationships between Ordered Compositions and Fibonacci Numbers

A sequence of four compositions of 3 is: 1 + 1 + 1, 1 + 2, 2 + 1, 3. By the replacement of the plus signs (+) and commas (,) by the multiplication dots (∙) and plus signs (+) respectively, the sequence becomes the summation series: 1∙1∙1 + 1∙2 + 2∙1 + 3, which is equal to 8 or 6 th number in the famous Fibonacci sequence. It is a curious fact that the sum of a positive integer n and the products of summands corresponding to the compositions of n is equal to (2n)-th Fibonacci number. We establish the proposition after obtaining a special order of the compositions of n; and then obtain some other results. The results show that Fibonacci sequence has close connection with the special order of the compositions of n. Two Fibonacci identities, which we derive from a special recurrence relation, are useful to prove two theorems. The relationships are stated first in the theorems and are then shown in the consequences of the theorems.


Compositions of n = C(n).
2. All C(n) = {C(n)}.Only for the mathematical representations, we use the notation: {C(n)} to mean all C(n).Otherwise we use an adjective to specify C(n).
Example: We use the symbol of equivalence (≡) between SOC(n) and its implication.

SOC(5) ≡
(b) By the replacement of the plus signs (+) and commas (,) by the multiplication dots (•) and plus signs (+) respectively in SOC(n), we can find a summation series of 2 n -1 terms.Denoting the summation series by ∏ , we show: =  2 such that the sum of the terms of odd places and the sum of the terms of even places of ∏ are  2 − 2 and  2 − 1 respectively.For example, when SOC(4 (ii) The 1 st term, the sums of 1 st 2, 1 st 2 2 , … 1 st 2  − 2 and all 2  − 1 terms of ∏ for n ≥ 2 are  2 ,  4 ,  6 , … ,  2 − 2 and  2 ; and the 2 nd term, the sums of 2 nd 2, 2 nd 2 2 , …, 2 nd 2  − 2 terms of the series are  3  5  2 − 1 in succession.
(c) Two Fibonacci identities: have the important roles to establish the relationships.We obtain the identities from a special recurrence relation.
In the description of the relationships, the basic results are stated first in two theorems; other results are the consequences of the theorems.

Ordered Compositions
By the notations, we can write for n ≥ 2, (1) is an initial sequential arrangement for all C(n) such that the arrangement is composed of n sets of C(n).The last two sets of C(n) are (n -1) + {C(1)} and n.Obviously these are two particular C(n) as: (n -1) + 1 and n respectively.When n = 2 then {C(2)} ≡ 1 + 1, 2; and when n ≥ 3 then besides the last two sets as two C(n), we can further obtain the sequential arrangements for other sets of C(n) of the type: a + {C(na)} where these arrangements have the same form as that of the initial sequential arrangement for all C(n).For instance, where C(n) in the 1 st set on the right have 1 st two common summands 2 and 1; C(n) in the 2 nd set have 1 st two common summands 2 and 2; and so on provided that the last two sets are two C(n) of 3 and 2 summands as: 2 + (n -3) + 1 and 2 + (n -2) respectively.Subsequently we can obtain the sequential arrangements for the sets of C(n) of the type: a + b + {C(nab)} in like manner.Thus carrying out the operations of obtaining the sequential arrangements for the successive sets of C(n) recursively where the last two sets in each arrangement are found as two C(n), finally we can find a definite order of all C(n).For instance, For convenience, we name 5 sets of C( 5), which is shown in the 1 st step, as the 'basic' or 'initial exposition' of all C(5); and name the sixteen C( 5), which are yielded in a definite order in the last step, as the 'final exposition' of all C(5).Thus the expression on the right of (1), which is composed of n sets of C(n), is the initial exposition of all C(n); and 2 n -1 C(n) in a particular order, which can be yielded finally by recursive exposition, is the final exposition of all C(n).The particular order of all C(n) is named as 'the significant order of compositions of n' or in brief SOC(n).SOC(n) is the final exposition of all C(n).
We can use also the phrases: 'initial exposition' and 'final exposition' in the expositions of a set of C(n).In the process of recursive exposition to find SOC(n) starting with (1), we find in general a set of C(n) in the form: x 1 + … + x r -1 + {C(x r )} for x 1 + … + x r = n such that the initial exposition of this set of C(n) is: Following (1), we can write: (1.1) is the initial exposition of 1 + {C(n -1)}.The exposition starts with 1 + 1 + {C(n -2)} of which the initial exposition (1.2) starts with 1 + 1 + 1 + {C(n -3)} of which the initial exposition (1.3) starts with 1 + 1 + 1 + 1 + {C(n -4)}; and so on.It follows that SOC(n) starts with the longest C(n) of which the summands are all 1. n as a C(n) is written at the last of (1).Consequently n as a C(n) appears last in SOC(n).Hence the sum of n summands, which are all 1, is the first C(n); and n itself is the last C(n) in SOC(n).For example, the first and last C(5) in SOC( 5) are + 1 + 1 + 1 + 1 and 5 respectively.Similarly x 1 + … + x r -1 + {C(x r )} for x 1 + … + x r = n, which represents a set of C(n), has the final exposition with the first C(n) as: x 1 + … + x r -1 + the sum of x r summands which are all 1 and the last C(n) as: In the process to find SOC(n), when a set of C(n) appears in the form:

Rule for SOC(n):
Under SOC(n), the summands of the 1 st C(n) are all 1; the last C(n) is n itself; and for n ≥ r ≥ 2, if any k th C(n) is: x 1 + … + x r then (k + 1) th C(n) is: x 1 + … + x r -2 + (x r -1 + 1) + the sum of x r -1 summands which are all 1 such that if r ≥ 3 then the first r -2 summands of k th C(n) appear also in (k + 1) th C(n) in the same order, but if r = 2 then such common summands of k th C(n) and (k + 1) th C(n) cannot exist.The number of summands of k th C(n) and (k + 1) th C(n) under SOC(n) are r and r + x r -2 respectively.
Remark1.The number of the compositions of n is a familiar result.It is easy to obtain the result from (1) also.
From (1), we get: Then we have the successive results as shown.
(2) can yield some more Fibonacci identities and also some Lucas identities for the particular values of the triplet (a, b, 0 ).

Two Theorems
Theorem 1.If ∏ for n ≥ 2 denotes the summation series such that ∏ = product of the summands of 1 st C(n) in SOC(n) + product of the summands of 2 nd C(n) in SOC(n) + ... + product of the summands of If the initial condition is defined as: ∏ 1 = 1 , then the result in general is: for n ≥ 1, ∏ =  2 .
Proof: It follows from (1) and the definition of ∏ that for n ≥ 2, The initial condition is: ∏ 1 = 1 =  2 .Then from (12), we get: We assume that the theorem is true for the first n natural numbers for any given n.Then from ( 12) and ( 8), we deduce that Hence we have the theorem by induction.▮ Theorem 2. If ∏ for n ≥ 2 denotes the summation series obtained by changing the same sign of the series ∏ for n ≥ 2 by alternating signs starting with + sign, then ∏ = − 2 − 3 .
In general for n ≥ 3, ∏ Now we follow the rules of induction.

Consequences of the Theorems
Theorem1 and Theorem 2 lead to find some relationships among ∏ , ∏ C and Fibonacci sequence.
From the consequences of the theorems, it follows that both ∏ C and ∏ have the sets of terms in the definite orders such that the sums of these sets of terms represent ordered Fibonacci numbers with some repetitions.Thus it is a remarkable fact that there exists a special order of the compositions of a natural number n, which has very close connection with the famous Fibonacci sequence.
then the next set of C(n) appears in the form: x 1 + … + x k + (a + 1) + {C(b -1)}.Let these two sets of C(n) be denoted by S 1 and S 2 respectively.Since S 1 and S 2 are the consecutive sets of C(n), it follows that the last C(n) in the final exposition of S 1 and the first C(n) in the final exposition of S 2 are two consecutiveC(n) under SOC(n).The last C(n) in the final exposition of S 1 is: x 1 + … + x k + a + b; and the first C(n) in the final exposition of S 2 is x 1 + … + x k + (a + 1) +the sum of b -1 summands which are all 1.Hence under SOC(n), when a C(n) appears in the form: x 1 + … + x k + a + b, then the next C(n) appears in the form: x 1 + … + x k + (a + 1) + the sum of b -1 summands which are all 1, where these two consecutive C(n) are composed of k + 2 and k + b summands respectively.The forms of S 1 and S 2 can be: a + {C(b)} and (a + 1) + {C(b -1)} respectively for a + b = n so that the last C(n) in the final exposition of S 1 is a + b and the first C(n) in the final exposition of S 2 is (a + 1) + the sum of b -1 summands which are all 1.Hence under SOC(n), the forms of two consecutive C(n) can be: (i) a + b and (ii) (a + 1) + the sum of b -1 summands which are all 1, where these two consecutive C(n) are composed of 2 and b summands respectively.The rule for appearances of successive C(n) under SOC(n) is clear from the above demonstration.