Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 1. pp. 149-160, 2011 Applied Mathematics
Some Algebraic Identities on the Integer Sequence Associated with Imaginary Quadratic Number Field Q(√−163)
Ahmet Tekcan
Uludag University, Faculty of Science, Department of Mathematics, Bursa,Turkiye e-mail: tekcan@ uludag.edu.tr
Received Date: November 24, 2010 Accepted Date: March 29, 2011
Abstract. In this work, we deduce some algebraic identities on the integer se-quence = ( ) with parameters and associated with the imaginary quadratic number field Q(√−163).
Key words: Fibonacci numbers; Lucas numbers; Pell numbers; integer se-quence; imaginary quadratic number field.
2000 Mathematics Subject Classification. 05A19, 11B37, 11B39. 1. Preliminaries
Fibonacci, Lucas and Pell numbers and their generalizations arise in the exam-ination of various areas of science and art. In fact these numbers are special case of a sequence which is defined as a linear combination as follows:
(1) += 1+−1+ 2+−2+ · · · +
where 1 2 · · · are real constants. The applications and identities related with these numbers can be seen in [1, 2, 3, 4, 5,9,10].
2. The Sequences( ) with Parameters and .
In [6, 7, 8], we deduced some algebraic relations on Lucas, Fibonacci and Pell numbers, respectively. In the present paper, we want to define a new integer sequence associated with the imaginary quadratic number field Q(√−163). To get this we first set let 0 be a square-free integer and set
∆ = (
4 ≡ 1 2( 4) ≡ 3( 4)
and call −∆ the discriminant of the imaginary quadratic number field K = Q(√−∆). Let ∆ denote the class number of K. Euler proved that the quadratic polynomials for which many values are primes. He observed that if is the prime 2 3 5 11 17 or 41, then the values (0) (1) · · · ( − 2) of the polynomial
(2) () = 2+ +
are prime (in other words, ( − 1) = 2 is not prime, so this sequence of consecutive prime values is the best one can hope for). For = 41, there are 40 prime numbers 41 43 47 53 · · · 1447 1523 1601. Also G. Rabinovitch (1912) proved that for prime , the integers (0) (1) · · · ( − 2) are all primes if and only if the imaginary quadratic field Q(√1 − 4) has class number 1. So an imaginary quadratic field Q(√1 − 4) has class number 1 if and only if 4 − 1 = 7 11 19 43 67 or 163, that is, = 2 3 5 11 17 or 41. Recall that a quadratic field K has class number 1 if every algebraic integer in K can be expressed as a product of primes in K and if any two such representations differ only by a unit, i.e. an algebraic integer that is a divisor in 1 in K. The imaginary quadratic fields of class number 1 were determined in 1966 by A. Baker and H. M. Stark, independently and free of the doubt that clung to Heegner’s earlier work in 1952. So = 41 is the largest prime number for which the values (0) (1) · · · ( − 2) of the polynomial in (2) are all primes.
In this section, our first aim is to define a new integer sequence associated with the class number of an imaginary quadratic field Q(√−163) and derive some algebraic identities on it. We see as above that for primes = 2 3 5 11 17 or 41, the values (0) (1) · · · ( − 2) of the polynomial () in (2) are prime. Also the integers (0) (1) · · · ( − 2) are all primes if and only if the imaginary quadratic field Q(√1 − 4) has class number 1, and = 41 is the largest prime number for which Q(√1 − 4) has class number 1. Now we take = 41. Then 41() = 2+ + 41. Let be any integer. Then
(3) 41() = 2+ + 41
We set = 0
41() = 2 + 1 and = 41() = 2+ + 41 Then we define the sequence = ( ) with parameters and as 0= 0, 1= 1 and (4) = −1− −2= (2 + 1)−1− (2+ + 41)−2
for all ≥ 2. The characteristic equation of (4) is 2−(2+1)+(2++41) = 0. The discriminant is hence = (2 + 1)2− 4(2+ + 41) = −163 So the roots of it are (5) = (2 + 1) + √ 163 2 = (2 + 1) − √163 2
Hence by Binet’s formula we get
(6) =
− −
for all ≥ 1. Now we can return our main problem. First, we give the following theorem concerning the sum of first non-zero numbers.
Theorem 2.1. Let denote the −th number. Then X =1 = (2+ + 41) − +1+ 1 2− + 41
Proof: Recall that = (2 + 1)−1− (2+ + 41)−2. So +2 = (2 + 1)+1− (2+ + 41) = 2+1+ +1− (2+ + 41) and hence
(7) +2− +1= 2+1− (2+ + 41) Applying (7), we deduce that
(8) = 0 ⇒ 2− 1= 21− (2+ + 41)0 = 1 ⇒ 3− 2= 22− (2+ + 41)1 = 2 ⇒ 4− 3= 23− (2+ + 41)2 · · · = − 1 ⇒ +1− = 2− (2+ + 41)−1 = ⇒ +2− +1= 2+1− (2+ + 41) If we sum both sides of (8), then we obtain
(9)
+2−1= [2−(2++41)](1+2+· · ·+)+2+1−(2++41)0 Since 0= 0 and 1= 1, (9) becomes +2− 1 = (−2+ − 41)(1+ 2+ · · · + ) + 2+1 and hence (10) 1+ 2+ · · · + = +2− 1 − 2+1 −2+ − 41 Taking +27→ (2 + 1)+1− (2+ + 41) in (10), we get 1+ 2+ · · · + = (2+ + 41) − +1+ 1 2− + 41 This completes the proof.
Now we can give the following theorem.
Theorem 2.2. Let denote the −th number. Then
+ = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ +1− (2+ + 41)−1 for ≥ 1 2+1− (2 + 1) for ≥ 0 (2 + 1)− 2(2+ + 41)−1 for ≥ 1
and + = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 2−1 P 2 =0 µ 2 ¶ (−1)(2 + 1)−2163 if is even 1 2−1 P−1 2 =0 µ 2 ¶ (−1)(2 + 1)−2163 if is odd for ≥ 1.
Proof: Note that +1= (2 + 1)− (2+ + 1)−1. Hence +1− (2+ + 41)−1 = (2 + 1)− 2(2+ + 41)−1 = (2 + 1) µ− − ¶ − 2(2+ + 41) µ−1− −1 − ¶ = (2 + 1) µ − − ¶ − 2(2+ + 41) µ − ( − ) ¶ = (2 + 1) √163 ( − ) − 2 √163( − ) = ∙ 2 + 1 − 2 √163 ¸ + ∙ −2 − 1 + 2 √163 ¸ = +
We see as above that +1− (2+ + 41)−1 = + . So (2+ + 41)−1= +1− (+ ) and hence + +1 = + [(2 + 1)− (2+ + 41)−1] = (2 + 2)− (2+ + 41)−1 = (2 + 2)− [+1− (+ )] = (2 + 2)− +1+ (+ ) So + = 2
+1− (2 + 1). Similarly it can be shown that + = (2 + 1) − 2(2 + + 41)−1 for ≥ 1 The last assertion is just an application to binomial series.
We can also give the −th number by using the powers of (2 + 1) and 163 or powers of (2 + 1) and (2+ + 41). To get this we can give the following theorem without giving its proof since it is just an application to binomial series. Theorem 2.3. Let denote the −th number. Then
1. = 1 2−1 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ −2 2 P =0 µ 2 + 1 ¶ (−1)(2 + 1)−(2+1)163 if is even −1 2 P =0 µ 2 + 1 ¶ (−1)(2 + 1)−(2+1)163 if is odd
2. = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ −2 2 P =0 µ − 1 − ¶ (−1)(2 + 1)−(2+1)(2+ + 41) if is even −1 2 P =0 µ − 1 − ¶ (−1)(2 + 1)−(2+1)(2+ + 41) if is odd. for all ≥ 1
Example 2.1.Let = 4. Then = 9−1 − 61−2. The first few numbers are 0 1 9 20 −369 −4541 −18360 111761 2125809 12314860 −18840609 · · · Let = 6. Then 6 = 1 25 2 X =0 µ 6 2 + 1 ¶ (−1)96−(2+1)163 = 1 25[6 · 9 5 − 20 · 93· 163 + 6 · 9 · 1632] = −18360
and let = 7, then
7 = 1 26 3 X =0 µ 7 2 + 1 ¶ (−1)97−(2+1)163 = 1 26[7 · 9 6 − 35 · 94· 163 + 21 · 92· 1632− 1633] = 111761 Similarly we get 6= 2 X =0 µ 5 − ¶ (−1)96−(2+1)61= 95− 4 · 93· 61 + 3 · 9 · 612= −18360 and 7 = 3 X =0 µ 6 − ¶ (−1)97−(2+1)61= 96− 5 · 94· 61 + 6 · 92· 612− 613 = 111761
Now we set the following identities = −2 2− 81 + √163 2√163 = − 2 + − 41 =2 + 3 + √ 163 2√163 =2 3+ 32− 241 − 121 + (32+ 3 − 39)√163 2√163 = 2 2+ 81 + √163 2(2+ + 41)√163 Then we can give the following theorem.
Theorem 2.4. Let denote the −th number. Then
1. The sum of first non-zero numbers is 1[ − − 1] 2. + +1= − for ≥ 0.
3. +1+ −1= −2− −2 for ≥ 2. 4. − −1= − for ≥ 1.
Proof:1. We proved in Theorem 2.2 that +=
+1−(2+ +41)−1. So +1+ +1 = +2− (2+ + 41) = (2 + 1)+1− 2(2+ + 41) = [+1− (2+ + 41)] + 2+1− (2+ + 41) and hence +1− (2+ + 41) = +1+ +1− 2+1+ (2+ + 41) = +1+ +1− 2 µ +1− +1 − ¶ + (2+ + 41) µ − − ¶ = µ − 2 √163 + 2+ + 41 √163 ¶ + µ + 2 √163 − 2+ + 41 √163 ¶ = Ã −22− 81 + √163 2√163 ! − Ã −22− 81 − √163 2√163 ! = −
2. Note that +1= (2 + 1)− (2+ + 41)−1. So +1+ = (2 + 2)− (2+ + 41)−1 = (2 + 2) µ − √163 ¶ − (2+ + 41) µ −1− −1 √163 ¶ = (2 + 2) µ − √163 ¶ − µ − √163 ¶ = µ 2 + 2 − √163 ¶ + µ −2 − 2 + √163 ¶ = Ã 2 + 3 + √163 2√163 ! − Ã 2 + 3 − √163 2√163 ! = −
3. We proved in Theorem 2.2 that + = 2
+1− (2 + 1). So + = 2+1− (2 + 1) = 2+1− (2 + 1)[(2 + 1)−1− (2+ + 41)−2] = 2+1− (2 + 1)2−1+ (2 + 1)(2+ + 41)−2 = 2(+1+ −1) − (42+ 4 + 3)−1+ (2 + 1)(2+ + 41)−2 and hence +1+ −1 = + + (42+ 4 + 3) −1− (2 + 1)(2+ + 41)−2 2 = + +42+4+3 √163 ( −1− −1) −(2+1)(2++41) √163 ( −2− −2) 2 = 2 ∙ 1 +4 2+ 4 + 3 √163 1 − (2 + 1)(2+ + 41) √163 1 2 ¸ + 2 ∙ 1 −4 2+ 4 + 3 √163 1 + (2 + 1)(2+ + 41) √163 1 2 ¸ = −2 " 23+ 32− 241 − 121 + (32+ 3 − 39)√163 2√163 # −−2 " 23+ 32− 241 − 121 − (32+ 3 − 39)√163 2√163 # = −2− −2
4. We proved in Theorem 2.2 that + = +1− (2+ + 41)−1. Hence +1= + + (2+ + 41)−1 Further we see that +1− (2+ + 41) = − . Hence +1 = − + (2+ + 41) So + + (2+ + 41) −1= − + (2+ + 41) and − −1 = + − + 2+ + 41 = (1 − ) + (1 + ) 2+ + 41 = ³1 −−22 −81+√163 2√163 ´ + ³1 + −22−81−√163 2√163 ´ 2+ + 41 = Ã 22+ 81 + √163 2(2+ + 41)√163 ! − Ã 22+ 81 − √163 2(2+ + 41)√163 ! = −
Now we can formulate the sum of even and odd numbers 2 and 2−1, respectively by using the powers of and .
Theorem 2.5. Let denote the −th number. Then
X =1 2= ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 2 P =1 4−3− 2 P =1 4−3 if is even 2 √163+ −1 2 P =1 4−3− 2 √163− −1 2 P =1 4−3 if is odd and X =1 2−1= ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 2 P =1 4−4− 2 P =1 4−4 if is even 2−1 √163 + −1 2 P =1 4−4− 2−1 √163 − −1 2 P =1 4−4 if is odd
Proof:We proved in (3) of Theorem 2.4 that +1+ −1= −2−−2 for ≥ 2. Now let be even. Then
X =1 2 = (2+ 4) + (6+ 8) + · · · + (2−2+ 2) = ( − ) + (5− 5) + · · · + (2−3− 2−3) = ( + 5+ · · · + 2−3) − ( + 5 + · · · + 2−3) = 2 X =1 4−3− 2 X =1 4−3
and let be odd, then X =1 2 = (2+ 4) + (6+ 8) + · · · + (2−4+ 2−2) + 2 = ( − ) + (5− 5) + · · · + (2−5− 2−5) + 2− 2 − = 2 √163+ ( + 5 + · · · + 2−5) − 2 √163 −( + 5+ · · · + 2−5) = 2 √163+ −1 2 X =1 4−3− 2 √163− −1 2 X =1 4−3
The other assertion can be proved similarly.
Now we want to derive a recurrence relation on numbers. To get this we can give the following theorem.
Theorem 2.6. Let denote the −th number. Then
2= (22+ 2 − 81)2−2− (4+ 23+ 832+ 82 + 1681)2−4 and
2+1= (22+ 2 − 81)2−1− (4+ 23+ 832+ 82 + 1681)2−3 for all ≥ 2
Proof: Note that 2 = (2 + 1)2−1− (2+ + 41)2−2 and hence 2 = (2 + 1)£(2 + 1)2−2− (2+ + 41)2−3¤− (2+ + 41)2−2 = 2−2[(2 + 1)2− (2+ + 41)] − (2 + 1)(2+ + 41)2−3 = 2−2[(2 + 1)2− (2+ + 41)] −(2 + 1)(2+ + 41)£(2 + 1)2−4− (2+ + 41)2−5¤ = 2−2[(2 + 1)2− (2+ + 41)] − (2 + 1)2(2+ + 41)2−4 +(2 + 1)(2+ + 41)22−5 = 2−2[(2 + 1)2− (2+ + 41)] − (2+ + 41)2−2 +(2+ + 41)2−2− (2 + 1)2(2+ + 41)2−4 +(2 + 1)(2+ + 41)22−5
= 2−2[(2 + 1)2− 2(2+ + 41)] + (2+ + 41)[(2 + 1)2−3 −(2+ + 41)2−4] − (2 + 1)2(2+ + 41)2−4 +(2 + 1)(2+ + 41)22−5 = 2−2[(2 + 1)2− 2(2+ + 41)] + (2 + 1)(2+ + 41)2−3 −(2+ + 41)22−4− (2 + 1)2(2+ + 41)2−4 +(2 + 1)(2+ + 41)22−5 = 2−2[(2 + 1)2− 2(2+ + 41)] +(2 + 1)(2+ + 41)[(2 + 1)2−4− (2+ + 41)2−5] −(2+ + 41)22−4− (2 + 1)2(2+ + 41)2−4 +(2 + 1)(2+ + 41)22−5 = 2−2[(2 + 1)2− 2(2+ + 41)] + (2 + 1)2(2+ + 41)2−4 −(2 + 1)(2+ + 41)22−5− (2+ + 41)22−4 −(2 + 1)2(2+ + 41)2−4+ (2 + 1)(2+ + 41)22−5 = 2−2[(2 + 1)2− 2(2+ + 41)] − (2+ + 41)22−4 = (22+ 2 − 81)2−2− (4+ 23+ 832+ 82 + 1681)2−4 The other assertion can be proved similarly.
For numbers, we set
= () = ∙ 2 + 1 −2− − 41 1 0 ¸ = () = ∙ 2 + 1 1 1 0 ¸ = () =£ 1 0 ¤ Then we can give the following result.
Theorem 2.8. Let denote the −th number. Then 1. = " +1 −(2+ + 41) −(2+ + 41)−1 # for ≥ 1 2. −1 = " +1 −1 # for ≥ 1
3. If is odd, then = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −1 2 P =0 µ − ¶ (2 + 1)−2 −1 2 P =0 µ − 1 − ¶ (2 + 1)−1−2 −1 2 P =0 µ − 1 − ¶ (2 + 1)−1−2 −3 2 P =0 µ − 2 − ¶ (2 + 1)−2−2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦
for ≥ 3 and if is even, then
= ⎡ ⎢ ⎢ ⎢ ⎣ 2 P =0 µ − ¶ (2 + 1)−2 −2 2 P =0 µ − 1 − ¶ (2 + 1)−1−2 −2 2 P =0 µ − 1 − ¶ (2 + 1)−1−2 −2 2 P =0 µ − 2 − ¶ (2 + 1)−2−2 ⎤ ⎥ ⎥ ⎥ ⎦ for ≥ 2 4. = +1 for ≥ 1
Proof: 1. We prove it by induction on . Let = 1. Then = " 2 −(2+ + 41)1 1 −(2+ + 41)0 # = " 2 + 1 −2− − 41 1 0 # So it is true for = 1. Let us assume that this relation is satisfied for − 1, that is, −1= " −(2+ + 41)−1 −1 −(2+ + 41)−2 # Since = · −1, we get = " 2 + 1 −2− − 41 1 0 # " −(2+ + 41)−1 −1 −(2+ + 41)−2 # = " +1 −(2+ + 41)[(2 + 1)−1− (2+ + 41)−2] −(2+ + 41)−1 # = " +1 −(2+ + 41) −(2+ + 41)−1 #
2. and 3. can be proved similarly. 4. = £ 1 0 ¤ " +1 −(2+ + 41) −(2+ + 41)−1 # ∙ 1 0 ¸ = £ 1 0 ¤ ∙ +1 ¸ = +1
3. Acknowledgements
This work was supported by the Commission of Scientific Research Projects of Uludag University, Project number UAP(F)-2010/55.
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