Solution of problem set 5
... where ak = Fk (mod n). Obviously, the series will begin (1,1). At some point in this series, (ai , ai+1 ) will equal (aj , aj+1 ). Since the Fibonacci numbers are cyclical, every ai is completely determined by ai−2 and ai1 . Therefore, the only way to get (1,1) is if the second number in the pair im ...
... where ak = Fk (mod n). Obviously, the series will begin (1,1). At some point in this series, (ai , ai+1 ) will equal (aj , aj+1 ). Since the Fibonacci numbers are cyclical, every ai is completely determined by ai−2 and ai1 . Therefore, the only way to get (1,1) is if the second number in the pair im ...
over Lesson 9–1
... The model shows 2 rows of 7 squares. The squares could also be arranged in 7 rows of 2 squares, 14 rows of 1 square, or 1 row of 14 squares, as shown below. ...
... The model shows 2 rows of 7 squares. The squares could also be arranged in 7 rows of 2 squares, 14 rows of 1 square, or 1 row of 14 squares, as shown below. ...
![[Part 2]](http://s1.studyres.com/store/data/008795912_1-134f24134532661a161532d09dceadfe-300x300.png)
Flat primes and thin primes
... Some interesting subclasses of primes have been identified and actively considered. These include Mersenne primes, Sophie Germain primes, Fermat primes, Cullen’s primes, Wieferich primes, primes of the form n2 + 1, of the form n! ± 1, etc. See for example [14, Chapter 5] and the references in that t ...
... Some interesting subclasses of primes have been identified and actively considered. These include Mersenne primes, Sophie Germain primes, Fermat primes, Cullen’s primes, Wieferich primes, primes of the form n2 + 1, of the form n! ± 1, etc. See for example [14, Chapter 5] and the references in that t ...
