INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES Spectrum of Fibonacci and Lucas numbers Asker Ali Abiyev Abstract It has been achieved polynomial function, depending on 3 3 2 n n a− b = a − b a + b− ab arguments a + b and ab arguments of expression a +b for the ()() biggest and smallest numbers which are in the centre of natural geometrical figures (line, square, cube,…,hypercube). The coefficients of this polynomial are defined from triangle tables, But there are no formulae for big values of n in literature. written by special algorithm by us. The sums of the numbers in The purpose of the article is to write new identities for the each row of the triangles make Lucas and Fibonacci sequences. binomial expression an± b n and study the fields of their New formulae for terms of these sequences have been suggested by application. us (Abiyev’s theorem). As the coefficients of the suggested polynomial are spectrum of Fibonacci and Lucas numbers they will opportunity these number’s application field to be enlarged. 2. F( a+b), ab EXPRESSION 2 Keywords: Lucas, Fibonacci, sequences, identity, Binet formula, If set of natural numbers {1, 2, 3, … , n}, {1, 2, 3, … , n }, 3 4 p polynomial. {1, 2, 3, … , n }, {1, 2, 3, … , n }, … , {1, 2, 3, … , n } is written accordingly in line, square, cube,…, hypercube of p 1.INTRODUCTION order one after another , the amount of numbers in their p centres is defined with q=2 formula. Let’s mark the n n maximal number with A and minimal number with B . The binomial expression a± b is one of the most used p p Where, n- even number and order of geometrical figure, p- expressions in algebra. It has evidently been seen up to now dimension of space. If p=3 and n=6, it is cube of 6 order in that one of the powers of a and b coefficients increases space of 3 dimension. and another one decreases in the identity of this expression The following formulae Ap(n) and Bp(n) achieved after [1]. The following formula can serve as an example. doing some algebraic operations by using numbers of figures’ centres for p=1, p=2, p=3, n=6 and n=8: abn− n =( aba −) n−1 + abab n − 2 + n − 3 2 +... + abn−2 + b n − 1 (1) ( ) n p−1 p − 2 p − 3 Ap () n= n +( n + n +... + 1) + 1 2 (2) When a and b letters are conjugate expressions, algebraic n calculus become more difficult by using (1) identity. As the p−1 p − 2 p − 3 Bp () n= n −( n + n +... + 1) (3) sum and multiplication of the conjugate expressions are real 2 numbers, writing the right side of (1) identity as a function depending on a+b and ab arguments simplifies the Let’s calculate A2(6), B2(6), A3(6), B3(6) by these formulae: calculus. This function can be easily written for small values 6 6 A ()6=[] 6 + 1 + 1 = 22 ; B ()6= 6 − 1 = 15 ; of n : 2 2 2 2 6 A 6= 62 + 6 + 1 + 1 = 3.43 + 1 = 130 3 3 2 3() () , a+ b =()() a + b a + b − 3 ab 2 6 B ()6= 62 −() 6 + 1 = 3.29 = 87 . 3 2 And now let’s find out sum and product of Ap(n) and Bp(n) : n p−1 p − 2 p − 3 n p−1 p − 2 p − 3 p Ap ()() n+ Bp n = n +( n +++++ n ... 1) 1 n −( n +++=+ n ... 1) n 1 2 2 (4) n n p n A n. B n= n p−1 +1 −n p−2 + n p − 3 +... + 1 n p−2 + n p − 3 +... + 1 + 1 p()() p ( ) ( ) 2 2 2 (5) By these formula let’s calculate A2(6) + B2(6), A2(6). B2(6) and A3(6)+ B3(6) , A3(6).B3(6): 3 A3(6) + B3(6)=6 +1=217=130+87; A3(6) B3(6)=3{36.109- (4) and (5) formulae have been proved by mathematical 7.[3.7+1]}=3.3770=11 310=130.87 induction method. Issue 1, Volume 6, 2012 165 INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES After using Binomial theorem and doing some algebraic operations, we get the following equalities for Ap+ 1 + B p + 1 = F A + B, A B expression: if p- odd p p ( p p) p p p+1 p + 1 1 p+1 Ap +B p= E p+1() A p + B p − A p B p × p+1 p+3 2 p−1 3 p−3 2 2 2 (6) ×EABABEABABEABEABp+1() p + p − p p p+1( p+ p ) − ... − p p p+1 ( p+ p ) − p+1 p p ... if p- even p+1 p + 1 Ap +B p 1 p =TABABp+1() p + p − p p × ABp+ p p p (7) +1 2p− 2 3 p − 42 2 2 TABABTABABTABTABp+1() p+ p − p p p + 1( p+ p ) − ... − p p p+1 ( p+ p ) − p + 1 p p ... p+3 p Table1, drawn with special algorithm by us [2]. Where +1 EEE1, 2 ,..., 2 TTT1, 2 ,..., 2 1 1 p+1 p + 1 p +1 and p+1 p + 1 p + 1 ETp+1 = p +1 =1 coefficients in (6) and (7) expressions are found from the Table 1. If to use (2) and (4) expressions, a new equality is found for ApBp product: p p s−1 s +1 n n n p−2 p − 3 n p−2 p − 3 ( p) ( p ) ABp p = +−1 (n +++ n .... 1) (n ++++= n .... 1) 1 −()AADp−1 −1 p−1 = p 2 2 2 2 4 (8) If to use (4) and (8) equalities, (6) and (7) will be as following: p+1 p+3 Ap+1 + Bp+1 = Es1p+ 1 − DEs 2p− 1 − DEs 3p− 3 −... − DEsED ( 2 2 − 2 )... p p p+1 p p p+1 p p p+1 p p p+1 p p+1 p (9) Issue 1, Volume 6, 2012 166 INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES p p +1 p+1 p+1 =+ 1 p − p−22 − ( p−43 −− (... 2 2 − 2 DTsTDsTDsTDsTsBA )..) p p +1 ppp +1 ppp +1 ppp pp +1 pp +1 p (10) 2 2 2 2 2 1 2 2 2 2 (n+ 2) n (n + 1) + 1 n n A1+=−=−=+− B 1 E 2 s 1 E 2 D 1 s 12 D 1 ( n 1) 2 −−= (AA0 1) 0 =++1 (11) 4 2 2 2 Let’s calculate (10) and (11) equalities for p=1, p=2 , p=3. n n Where A =1, p=1, A = +1 , B1 = . 0 1 2 2 For p=2 3 3 1 2 2 2 2 ()()s2 +1 s2 −1 (12) ABsTsTD2+ 223232 =( −)() = ss 22 −3 D 2 = ss 22 − 3 −()AA1 −1 1 = 4 n2+ 2 n 2 2 2 2 () n n =()()n +1 n +1 − 3 − +1 4 2 2 For p=3 n3 + 2 n 3 4 4 3 4 ( ) n n A3+ B 3 =() n +1 − −()n + 1 () n +1 + 1 × 4 2 2 n3 + 2 n 3 3 2 () n n ×4 ()n +1 − 2 −()n +1 () n +1 + 1 4 2 2 (13) 3. ABIYEV TRIANGLES As it seen from the formulae, as dimension of space It should be mentioned that, formulae (6) and (7), which increases, (12) and (13) equalities become more complicated suggested for natural numbers in centres of geometric depending on order of geometrical figure. Only p=1, figures, are true as well as for any numbers. Pythagorean theorem is satisfied. Because, the orders of If p+1=n, Ap=a and Bp=b in (6) and (7) equalities and expressions are the same in both right and left of equality open the brackets, they will be written as follows: [3]. n−1 n−3 n-12 n- 3 n+ 1 n-1 abn+ n = abTab +1 + − Tab2 + ab +... + Taby2 + 2 − Tab 2 2 ()( n ()()()n n () n () ) (7’) n n The identity a± b = F( a+b), ab was side and figure 1 to the left side of this column; write figure 2 before even numbers at the right and figure 0 at the left. In first revealed and a special table was compiled to determine consequence, 2 columns formed at the right and at the left of the coefficients in this identity [4]. Let’s explain the column 0. algorithm first to write the table: insert columns to the right nd rd Afterwards, write figure 1 in the 2 and 3 rows of the and left of 0 at the top of the table 1 and write numbers nd nd 2 columns; by summing the figure 1 in the 2 row of the beginning from 1 in sequence in column 0. Write odd nd rd st 2 column with the figure 3 in the 3 row of the 1 column numbers to the right Issue 1, Volume 6, 2012 167 INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES write figure 4 in the 4th row of the 2nd column. So by If the method used to form the columns at the right side continuing this method figures 4, 9, 16, 25 etc. squares of of the table1is applied symmetrically with respect to column sequent numbers are written at the right in the even rows of 0, columns can be written at the left side too.