A New 3n − 1 Conjecture Akin to Collatz Conjecture
... continue this process recursively. The 3n − 1 conjecture is that this process which will eventually reach either 1 or 5 or 17, regardless of which positive integer is selected at the beginning. The smallest i such that ai = 1 or 5 or 17 is called as the total stopping time of n. The 3n−1 conjecture ...
... continue this process recursively. The 3n − 1 conjecture is that this process which will eventually reach either 1 or 5 or 17, regardless of which positive integer is selected at the beginning. The smallest i such that ai = 1 or 5 or 17 is called as the total stopping time of n. The 3n−1 conjecture ...
THE CHINESE REMAINDER THEOREM INTRODUCED IN A
... and a man driving a horse hit the basket and broke all the eggs. Wishing to pay for the damage, he asked the girl how many eggs there were. The girl said she did not know, but she remembered that when she counted them by twos, there was one left over; when she counted them by threes, there were two ...
... and a man driving a horse hit the basket and broke all the eggs. Wishing to pay for the damage, he asked the girl how many eggs there were. The girl said she did not know, but she remembered that when she counted them by twos, there was one left over; when she counted them by threes, there were two ...
34(2)
... where m + n + p + q = 1 (mod 2) and (m, n, p, q) = 1. In the formulas above, either three or four variables are needed to generate four other integers (a, b,c,d). In this paper, we present 2-parameter Pythagorean quadruple formulas where the two integral parameters are also part of the solution set. ...
... where m + n + p + q = 1 (mod 2) and (m, n, p, q) = 1. In the formulas above, either three or four variables are needed to generate four other integers (a, b,c,d). In this paper, we present 2-parameter Pythagorean quadruple formulas where the two integral parameters are also part of the solution set. ...
Powers of Two as Sums of Two Lucas Numbers
... Let (Fn )n≥0 be the Fibonacci sequence given by F0 = 0, F1 = 1 and Fn+2 = Fn+1 + Fn for all n ≥ 0. The Fibonacci numbers are famous for possessing wonderful and amazing properties. They are accompanied by the sequence of Lucas numbers, which is as important as the Fibonacci sequence. The Lucas seque ...
... Let (Fn )n≥0 be the Fibonacci sequence given by F0 = 0, F1 = 1 and Fn+2 = Fn+1 + Fn for all n ≥ 0. The Fibonacci numbers are famous for possessing wonderful and amazing properties. They are accompanied by the sequence of Lucas numbers, which is as important as the Fibonacci sequence. The Lucas seque ...
http://waikato.researchgateway.ac.nz/ Research Commons at the
... corresponding even perfect number, (see [28]). By observing the form of Mersenne numbers, we can see that a higher Mersenne prime can come from a lower Mersenne prime. For example, 7 is a Mersenne prime, and a new Mersenne prime 127 can be obtained from 27 − 1. It was hoped that if the number Mn was ...
... corresponding even perfect number, (see [28]). By observing the form of Mersenne numbers, we can see that a higher Mersenne prime can come from a lower Mersenne prime. For example, 7 is a Mersenne prime, and a new Mersenne prime 127 can be obtained from 27 − 1. It was hoped that if the number Mn was ...
MIDTERM 1 TUESDAY, FEB 23 SOLUTIONS 1.– (15 points
... Euler’s theorem by simply squaring our result, while there is no obvious way to deduce our result form Euler’s theorem. It is not possible to improve further, and get an even smaller exponent, by question c. RemarkL” Hence the smallest natural number u such that au ≡ 1 (mod 2n+2 ) for all odd a is 2 ...
... Euler’s theorem by simply squaring our result, while there is no obvious way to deduce our result form Euler’s theorem. It is not possible to improve further, and get an even smaller exponent, by question c. RemarkL” Hence the smallest natural number u such that au ≡ 1 (mod 2n+2 ) for all odd a is 2 ...
Primitive Roots Modulo Primes - Department of Mathematics
... For positive integer n>1 and integer x, if there exists a least positive integer d such that xd ≡1 (mod n), then we say d is the order of x (mod n). We denote this by ordn(x) = d. It is natural to ask for a prime p, if there exists x such that ordp(x) = p−1. Such x is called a primitive root (mod p) ...
... For positive integer n>1 and integer x, if there exists a least positive integer d such that xd ≡1 (mod n), then we say d is the order of x (mod n). We denote this by ordn(x) = d. It is natural to ask for a prime p, if there exists x such that ordp(x) = p−1. Such x is called a primitive root (mod p) ...
The Farey Sequence and Its Niche(s)
... In the third he gives an example of F5 . In the final paragraph Farey writes, “I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of some easy or general demonstration?; which are points on which I should be glad to learn the ...
... In the third he gives an example of F5 . In the final paragraph Farey writes, “I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of some easy or general demonstration?; which are points on which I should be glad to learn the ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".