ON AN EXTENSION OF FIBONACCI SEQUENCE

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1 Bulletin of the Marathwada Mathematical Society Vol.7, No., June 06, Pages 8. ON AN EXTENSION OF FIBONACCI SEQUENCE S. Arolkar Department of Mathematics, D.M. s College and Research Centre, Assagao, Bardez, GOA suchita.golatkar@yahoo.com and Y.S.Valaulikar, Department of Mathematics, Goa University, GOA ysv@unigoa.ac.in Abstract Fibonacci sequence is extended in such a way that the coefficients of the terms, on the right hand side of its recurrence relation, are terms of the binomial expansion. Properties of two such extensions namely second and third order recurrence relations are discussed. INTRODUCTION The well known classical Fibonacci sequence is defined by F n+ = F n + F n, for all n with F 0 = 0 and F =, (.) where F n is the n th Fibonacci number. The classical Fibonacci sequence has been extended in many ways [6]. One way of generalizing this sequence as given in [3] is F n+ = a F n + b F n, for all n with F 0 = 0 and F =, (.) where a and b are any non zero real numbers. With a = k, b =, we get equation () of []. In the present paper, we look at the equation (.) from another point of view. We consider the coefficients on the right hand side of (.), namely a and b to be the terms of the Binomial expansion of (a + b). We will call it as B- Fibonacci sequence and write it as ( f B) n+ = a ( f B) n + b ( f B) n, for all n with ( f B) 0 = 0 and ( f B) =, Marathwada Mathematical Society, Aurangabad, India, ISSN ,

2 S.Arolkar AND Y.S.VALAULIKAR where ( f B) n is n th term of B-Fibonacci sequence (.3). The change in notation from F to ( f B) is made with expected further extensions. For example, the B-Tribonacci sequence will be defined by ( t B) n+ = a ( t B) n+ + ab ( t B) n + b ( t B) n, for all n with ( t B) 0 = 0, ( t B) = 0, and ( t B) =, (.4) where the coefficients on the right hand side are the terms of the binomial expansion of (a + b). It is well known that the binomial coefficients carry a lot of combinatorial information in them. As Binomial expansion is an important tool in Combinatorics related fields, it is natural to expect some applications of such sequences. The first few terms of sequence (.3) are ( f B) 0 = 0, ( f B) =, ( f B) = a, ( f B) 3 = a + b, ( f B) 4 = a 3 + ab, ( f B) 5 = a 4 + 3a b + b, ( f B) 6 = a 5 + 4a 3 b + 3ab and so on. Rewriting equation (.3),we get, ( f B) n = b ((f B) n+ a ( f B) n ), for all n with ( f B) 0 = 0, and ( f B) =. The sequence defined by equation (.5) is called Nega -B-Fibonacci sequence. Note that ( f B) = b, (f B) = a, ( f B) b 3 = a +b and so on. b 3 Following identities can be derived from equations (.3) and (.5) () The n th term of (.3) and also of (.5), is given by ϕ n ϕ ϕ + ϕn ϕ ϕ, a + 4b 0, ( f B) n = nϕ n, a + 4b = 0, ϕ = ϕ = ϕ. (.5) (.6) where ϕ = a+ a +4b and ϕ = a a +4b, for all a, b R \ {0}, are roots of the characteristics equation of (.3). Equation (.6) is a Binet type formula for (.3) and (.5). () (a) The generating function for B-Fibonacci sequence (.3) is given by, G (x) = x(a+bx) and that of (.5) is G ( ) (x) = x(a+bx) x (a+bx). (3) (a)the n th term of (.3) is also given by, ( f B) n = [ n ] (n r) r r! a n r b r, for all n. (.7) where (n r) r is n r to the r falling factorial.[6] (b)the n th term of (.5) is given by, ( f B) n = [ n ] ( ) n+ (n r) r b n r! a n r b r, for all n. (.8)

3 ON AN EXTENSION OF FIBONACCI SEQUENCE 3 (4) (a) The sum of the first (n + ) terms of (.3) is given by, n ( f B) r = b(f B) n + ( f B) n+, (.9) a + b provided a + b. (b)the sum of the first n terms of (.5), is given by, for n, provided for n, provided n ( f B) r = b(f B) (n+) + ( f B) n, (.0) a + b r= n r= a + b. Note that equation (.0) can be also written as ( f B) r = b(f B) (n+) + ( f B) n, (.) a + b a + b. Combining (.9) and (.), we have n ( f B) r = b((f B) n ( f B) (n+) ) + (( f B) n+ ( f B) n ), (.) a + b r= n Remarks: () Suppose a + b = and b =, then B-Fibonacci equation (.3) reduces to ( f B) n+ = ( f B) n ( f B) n and its r th term is ( f B) r = r. Hence n (f B) r = n(n+). () If a = 0 and b =, then sequence ( f B) n = {0, }. (3) As n becomes large and large, the n th term ( f B) n of (.3), will approach +b, whenever b < and that of (.5) will approach to +b, whenever b >. (5) In Matrix form the B-Fibonacci sequence (.3) (and also (.5)) is represented by [ ( f ] [ ] [ B) n 0 ( ( f = f ] B) n B) n+ b a ( f. B) [ ] [ n 0 ( Let A= = f B) 0 ( f ] [ B) b ( b a b( f B) ( f, then A B) n = f B) n ( f ] B) n b ( f B) n ( f. B) n+ Following identities can be proved by using the above matrix representation. (6) (a) ( f B) n ( f B) n ( f B) n+ ( f B) n = ( b)n, for all integer n. (b) Honsbergers type identity ( f B) n+m = b( f B) n ( f B) m + ( f B) n ( f B) m. With m = n identity (6b) reduces to

4 4 S.Arolkar AND Y.S.VALAULIKAR (c) ( f B) n = b ( f B) n + (f B) n. With m = n + identity (6b) reduces to (d) ( f B) n = b( f B) n ( f B) n + ( f B) n ( f B) n+. Also using (.3) and (6d),we can obtain, ( f B) n = a( f B) n + b ( f B) n ( f B) n. (7) General bilinear formula For any integers m, m, n and n, with m + n = m + n, (f B) m ( f B) n ( f B) m ( f B) n = ( f B) ( b)s m s ( f B) n s ( f B) m s ( f B) n s. (8) Catalan type identity For any integers n, r, ( f B) n ( f B) n+r ( f B) n r ( f B) n = ( b)n ( f B) r ( f B) r = ( b) n r ( f B) r. (9) d, Ocagne, type identity For any integers n,m, ( f B) n+ ( f B) m ( f B) n ( f B) m+ = ( b) n ( f B) m n, i.e. ( f B) m ( f B) n ( f B) m+ ( f B) n+ = ( b)n ( f B) m n. (0) ( f B) n ( f B) n ( f B) n+r ( f B) n+r = ( b)n ( f B) r, for all integers n and r. THE B-TRIBONACCI SEQUENCE In this section, we deal with B-Tribonacci sequence given by the third order recurrence relation (.4). The first few terms of the sequence are ( t B) 0 = 0, ( t B) = 0, ( t B) =, ( t B) 3 = a, ( t B) 4 = a 4 + ab and ( t B) 5 = a 6 + 4a 3 b + b. Rewriting equation (.4), we get ( t B) n = b ((t B) n+ a ( t B) n+ ab ( t B) n ), for all n with ( t B) 0 = 0, ( t B) = 0, and ( t B) =. (.) The sequence defined by equation (.) is called Nega -B-Tribonacci sequence. The first few terms of Nega B-Tribonacci sequence are ( t B) =, ( t B) b = a, b 3 ( t B) 3 = 3a, ( t B) b 4 4 = 4a3 + = ( 4a 3 b + b ) and so on. b 5 b 4 b 6 We have following identities for B-Tribonacci sequence.

5 ON AN EXTENSION OF FIBONACCI SEQUENCE 5 () The n th term of B-Tribonacci sequence (.4) (and also of (.)) is given by ϕ n (ϕ ϕ )(ϕ ϕ 3 ) + ϕ n (ϕ ϕ )(ϕ ϕ 3 ) + ϕ n 3 (ϕ 3 ϕ )(ϕ 3 ϕ ), ϕ i s are all distinct ( t B) n = ϕ n (ϕ ϕ ϕn ) (ϕ ϕ + nϕn ) (ϕ ϕ ), ϕ i s are such that ϕ ϕ = ϕ 3. (.) for all a, b R \ {0} and ϕ, ϕ and ϕ 3 are roots of the characteristic equation λ 3 a λ abλ b = 0 corresponding to (.4) and also of (.). The case of all three roots of the characteristic equation being equal, is ruled out due to choice of the coefficients. Equation (.) is a Binet type formula for (.4) and (.). () The generating function for B-Tribonacci sequence (.4) is given by ( t G) (x) = and that of (.) is t G x(a+bx) (x) = ( b b (x3 +b (ax+b) ) (3) (a) The n th term of the B-Tribonacci sequence (.4) is [ ] n 3 3 ( t (n 4 r) r B) n = a n 4 3r b r, n 3. (.3) r! where (n r) r is n r to the r falling factorial. (b) The n th term of (.) is given by ) ( t B) n = [ n ] ( ) n++r (n r) r+ b n+ (r + )! a n 3r b r, n. (.4) (4) (a)the sum of the first n + terms of (.5) is n+ ( t B) r = b ( t B) n + (b + ab)( t B) n+ + ( t B) n+ (a + b), (.5) provided a + b,. (b)the sum of the first n + terms of (.) is n ( t B) r = b ( t B) (n+) + (b + ab)( t B) (n+) + ( t B) n (a + b), (.6) provided a + b,. Rewriting (.6), we have n ( t B) r = b ( t B) (n+) + (b + ab)( t B) (n+) + ( t B) n (a + b), (.7) provided a + b,. Combining (.5) and (.7), we have [ n+ r= n (t B) r = (a+b) ] b (( t B) n ( t B) (n+) )+(b +ab)(( t B) n+ ( t B) (n+) )+(( t B) n+ ( t B) n ) (.8)

6 6 S.Arolkar AND Y.S.VALAULIKAR provided a + b,. We now state the excluded cases. (5) Let a, b R. (a) If a + b =, the r th term of B- Tribonacci sequence (.4) is given by ( t B) r = r p=0 ( b)p + [ r 3 ] r 4 3p p=0 s=r p ( )s (r 4 p) s+p p!s! b s+p. Hence we have, n+ r= (t B) r = n+ r r= p=0 ( b)p + n+ r= [ r 3 ] r 4 3p p=0 s=r p ( )s (r 4 p) s+p p!s! b s+p [ r 3 ] p=0 = ( b)n +b (n +b +b )+ n+ r= provided b. If b =, then a = 3 r 4 3p s=r p ( )s (r 4 p) s+p and equation (.4) reduces to p!s! b s+p, ( t B) n+ = 9 4 (t B) n+ 3 (t B) n + 4 (t B) n. (.9) The characteristics roots of (.9) are ϕ = 4, ϕ = = ϕ 3 and Binet s formula gives the r th term ( t B) r = n+ (t B) r = n+ = ( 4 3 )[ 4 3 ϕ r (ϕ ϕ ϕr ) (ϕ ϕ + rϕr ) ϕ ϕ ( ( 4 )n+) (n + ) ] + 3 ϕ r (ϕ ϕ ϕr ) (ϕ ϕ + rϕr ) (n + )(n + ). ϕ ϕ. Hence, (b) If a + b =, then the n th term of B- Tribonacci sequence (.4) is given by ( t B) n = n (b)r + [ n 3 ] ( )r n 4 3r (n 4 r) s+r s=n r r!s! b s+r This case can also be discussed as above. (6) The representation of B-Tribonacci sequence (.4) (and also Nega B-Tribonacci sequence (.)) in Matrix form is ( t B) n 0 0 ( t B) n ( B) n+ = 0 0 ( t B) n. ( t B) n+ b ab a ( t B) n+ 0 0 ( t B) 0 ( t B) ( t B) Let A= 0 0 = ( t B) ( t B) 0 ( t B), b ab a b ( t B) n b ( t B) ab( t B) ( t B) 3 b ( t B) n + ab( t B) n ( t B) n then A n = b ( t B) n b ( t B) n + ab( t B) n ( t B) n+. b ( t B) n+ b ( t B) n + ab( t B) n+ ( t B) n+ Following identities can be easily proved by using the above matrix representation. (7) (a) ( t B) n ( t B) n ( t B) n ( t B) n+ ( t B) n ( t B) n ( t B) n+ ( t B) n+ ( t B) n = [( b) ] n, for all integer n.

7 ON AN EXTENSION OF FIBONACCI SEQUENCE 7 (b) ( t B) n+m = b ( t B) n ( t B) m + ( t B) m (b ( t B) n + ab( t B) n ) + ( t B) n ( t B) m+. In particular, if m = n, then we have, (c) ( t B) n = b ( t B) n + (t B) n ( t B) n+ a ( t B) n, and with m = n +, identity 6(b) yields, (d) ( t B) n = ( t B) n+ + ab(t B) n + b ( t B) n ( t B) n. (8) General Trilinear identity For all integers m i, n i, r i, i 3, ( t B) m ( t B) n ( t B) r ( t B) m ( t B) n ( t B) r ( t B) m3 ( t B) n3 ( t B) r3 = [( b) ] s ( t B) m s ( t B) n s ( t B) r s ( t B) m s ( t B) n s ( t B) r s ( t B) m3 s ( t B) n3 s ( t B) r3 s provided n i + r k = n k + r i, m i + n j = m j + n i, m j + r k = m k + r j, for distinct i, j, k such that i, j, k =,, 3. Following identities can be deduced from identity (8). (9) Catalan type identity For any integers n,r, ( t B) n ( t B) n+r ( t B) n+r ( t B) n r ( t B) n ( t B) n+r ( t B) n r ( t B) n r ( t B) n = [( b) ] n (( t B) r( t B) r + ( t B) r( t B) r ). (0) d, Ocagne, type identity For any integers m,n and r, ( t B) m ( t B) n ( t B) r ( t B) m+ ( t B) n+ ( t B) r+ ( t B) m+ ( t B) n+ ( t B) r+ = [( b) ] r (( t B) m r ( t B) n r+ ( t B) m r+ ( t B) n r ). () For any integers n,r, ( t B) n ( t B) n ( t B) n ( t B) n+ ( t B) n ( t B) n ( t B) n+r ( t B) n+r ( t B) n+r = [( b) ] n ( t B) r. Remarks: All the identities in this paper can be proved using induction or Binet type formula.

8 8 S.Arolkar AND Y.S.VALAULIKAR References [] D.Burton, Elementary Number Theory, 6th edition, Tata McGraw-Hill, 006. [] S.Falcón,Á. Plaza,On k-fibonacci sequences and Polynomials and their derivatives, ScienceDirect, Chaos and Fractrals 39(009) [3] D.Kalman, R. Mena, The Fibonacci Numbers- Exposed VOL. 76, NO. 3, JUNE [4] W.G.Kelley, A.C.Peterson, Difference equations: an introduction with applications, Academic Press, An imprint of Elsevier,Second edition. [5] A. G. Shannon, A.F. Horadam, A Generalized pythagorean theorem,fibonacci Quarterly, Vol. 9, No. 3, 97, pp [6] S.Vajda, Fibonacci and Lucas numbers and the Golden section: Theory and Applications, Dover Publications,008.

THE GENERALIZED TRIBONACCI NUMBERS WITH NEGATIVE SUBSCRIPTS

#A3 INTEGERS 14 (014) THE GENERALIZED TRIBONACCI NUMBERS WITH NEGATIVE SUBSCRIPTS Kantaphon Kuhapatanakul 1 Dept. of Mathematics, Faculty of Science, Kasetsart University, Bangkok, Thailand fscikpkk@ku.ac.th