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- akdevaraj | planetmath.org

Failure functions Polynomial structure editors World Writable My contributed problems Shantha primes My questions E Galois Forename akdevaraj Surname City State Country Homepage ok Preamble this is the default PlanetMath preamble as your knowledge of TeX increases you will probably want to edit this but it should be fine as is for beginners almost certainly you want these usepackage amssymb usepackage amsmath usepackage amsfonts used for TeXing text within eps

Original URL path: http://www.planetmath.org/users/akdevaraj (2016-04-25)

Open archived version from archive - Carmichal number 561 | planetmath.org

bases to which 561 is a pseudoprime I found it is pseudoprime to bases 12 i 22 i 34 i and 56 i Needless to say it is a pseudoprime to other bases too More on this later Type of Math Object Definition Major Section Reference Add a correction Attach a problem Ask a question Search form Search Home Articles Questions Forums Planetary Bugs HS Secondary University Tertiary Graduate Advanced

Original URL path: http://www.planetmath.org/carmichalnumber5611 (2016-04-25)

Open archived version from archive - Carmichal number 561 | planetmath.org

can use pari to find the bases to which 561 is a pseudoprime I found it is pseudoprime to bases 12 i 22 i 33 i and 44 i Needless to say it is a pseudoprime to other bases too More on this later Type of Math Object Definition Major Section Reference Add a correction Attach a problem Ask a question Search form Search Home Articles Questions Forums Planetary Bugs

Original URL path: http://www.planetmath.org/carmichalnumber561-0 (2016-04-25)

Open archived version from archive - Carmichal number 561 | planetmath.org

can use pari to find the bases to which 561 is a pseudoprime I found it is pseudoprime to bases 12 i 22 i 33 i and 44 i Needless to say it is a pseudoprime to other bases too More on this later Type of Math Object Definition Major Section Reference Add a correction Attach a problem Ask a question Search form Search Home Articles Questions Forums Planetary Bugs

Original URL path: http://www.planetmath.org/carmichalnumber561-1 (2016-04-25)

Open archived version from archive - Carmichal number 561 | planetmath.org

found it is pseudoprime to bases 12 i 22 i 33 i and 44 i Needless to say it is a pseudoprime to other bases too More on this later Type of Math Object Definition Major Section Reference Add a correction Attach a problem Ask a question Comments Fallout of amateur research Permalink Submitted by akdevaraj on Tue 08 04 2015 04 39 We can use pari to find the smallest divisor of phi n for which a congruence holds good Example consider the pseudoprime 341 This is pseudo to the base 2 To find d the smallest divisor we run the program p n 2 n 1 341 It was found that when n 10 the function is exactly divisible by 341 Application When the program p n 2 n 97 341 was run for n 1 10 no divisibility was found Hence 2 n 97 is not divisible for any value of n However 2 n 1007 is divisible by 341 before the program reaches 10 Similarly 40 is the smallest divisor of phi 561 561 is a Carmichael number Hence any relevant program pertaining to an exponential expression has to be run only till n reaches 40

Original URL path: http://www.planetmath.org/carmichalnumber561 (2016-04-25)

Open archived version from archive - Failure functions | planetmath.org

on Wed 05 27 2015 06 06 Background see messages Abstract definition let p h i x p h i x phi x be a function of x x x Then x p s i x 0 x p s i subscript x 0 x psi x 0 is a failure if p h i p s i x 0 p h i p s i subscript x 0 phi psi x 0 is a failure in accordance with our definition of a failure Examples 1 Let our definition of a failure be a composite number Let the parent function ϕ x ϕ x phi x be a polynomial ring where the variable and coefficients belong to Z Z Z Then x p s i x 0 x 0 k ψ x p s i subscript x 0 subscript x 0 k ψ x psi x 0 x 0 k psi is a failure function since ϕ p s i x 0 ϕ p s i subscript x 0 phi psi x 0 will generate only failures composites 2 Let our definition of a failure again be a composite number Let the parent function ϕ x ϕ x phi x be an exponential function Then x x x ψ x 0 x 0 E u l e r p h i ϕ x 0 ψ subscript x 0 subscript x 0 E u l e r p h i ϕ subscript x 0 psi x 0 x 0 Eulerphi phi x 0 is a failure function since the parent function will now generate only failures composites 3 Let our definition of a failure be a non Carmichael number Let the parent function be 2 n 49 superscript 2 n 49 2 n 49 Then n 5 6 k n 5 6 k n 5 6 k is a failure function Here k k k belongs to N N N Applications a fragments a normal a indirect primality testing and b fragments b normal b in proving theorems in number theory Log in to post comments failure functions Permalink Submitted by akdevaraj on Wed 05 27 2015 06 06 Background see messages Abstract definition let p h i x p h i x phi x be a function of x x x Then x p s i x 0 x p s i subscript x 0 x psi x 0 is a failure if p h i p s i x 0 p h i p s i subscript x 0 phi psi x 0 is a failure in accordance with our definition of a failure Examples 1 Let our definition of a failure be a composite number Let the parent function ϕ x ϕ x phi x be a polynomial ring where the variable and coefficients belong to Z Z Z Then x p s i x 0 x 0 k ψ x p s i subscript x 0 subscript x 0 k ψ x psi x 0 x 0 k psi is a failure function since ϕ p s i x 0 ϕ p s i subscript x 0 phi psi x 0 will generate only failures composites 2 Let our definition of a failure again be a composite number Let the parent function ϕ x ϕ x phi x be an exponential function Then x x x ψ x 0 x 0 E u l e r p h i ϕ x 0 ψ subscript x 0 subscript x 0 E u l e r p h i ϕ subscript x 0 psi x 0 x 0 Eulerphi phi x 0 is a failure function since the parent function will now generate only failures composites 3 Let our definition of a failure be a non Carmichael number Let the parent function be 2 n 49 superscript 2 n 49 2 n 49 Then n 5 6 k n 5 6 k n 5 6 k is a failure function Here k k k belongs to N N N Applications a fragments a normal a indirect primality testing and b fragments b normal b in proving theorems in number theory Log in to post comments failure functions Permalink Submitted by akdevaraj on Wed 05 27 2015 06 06 Background see messages Abstract definition let p h i x p h i x phi x be a function of x x x Then x p s i x 0 x p s i subscript x 0 x psi x 0 is a failure if p h i p s i x 0 p h i p s i subscript x 0 phi psi x 0 is a failure in accordance with our definition of a failure Examples 1 Let our definition of a failure be a composite number Let the parent function ϕ x ϕ x phi x be a polynomial ring where the variable and coefficients belong to Z Z Z Then x p s i x 0 x 0 k ψ x p s i subscript x 0 subscript x 0 k ψ x psi x 0 x 0 k psi is a failure function since ϕ p s i x 0 ϕ p s i subscript x 0 phi psi x 0 will generate only failures composites 2 Let our definition of a failure again be a composite number Let the parent function ϕ x ϕ x phi x be an exponential function Then x x x ψ x 0 x 0 E u l e r p h i ϕ x 0 ψ subscript x 0 subscript x 0 E u l e r p h i ϕ subscript x 0 psi x 0 x 0 Eulerphi phi x 0 is a failure function since the parent function will now generate only failures composites 3 Let our definition of a failure be a non Carmichael number Let the parent function be 2 n 49 superscript 2 n 49 2 n 49 Then n 5 6 k

Original URL path: http://www.planetmath.org/failurefunctions-0 (2016-04-25)

Open archived version from archive - induction proof of fundamental theorem of arithmetic | planetmath.org

end cases is a positive integer less than n n n Since p n 0 fragments p normal subscript n 0 p mid n 0 the induction hypothesis implies that the prime p p p is in the prime decomposition of q p c q p c q p c and thus also at least of q p q p q p or c c c But we know that p c not divides p c p nmid c whence p q p fragments p normal q p p mid q p Thus we would get p q p p q fragments p normal q p p q p mid q p p q Because both p p p and q q q are primes it would follow that p q p q p q This contradicts the fact that p q p q p q Consequently our antithesis is wrong Accordingly n n n has only one prime decomposition and the induction proof is complete References 1 Esa V Vesalainen Zermelo ja aritmetiikan peruslause Solmu 1 2014 2 Ernst Zermelo Elementare Betrachtungen zur Theorie der Primzahlen Wissenschaftliche Gesellschaft zu Göttingen 1934 English translation in 3 H D Ebbinghaus A Kanamori eds Ernst Zermelo Collected Works Volume I Set Theory Miscellanea Springer 2010 Ernst Zermelo Elementary considerations concerning the theory of prime numbers 576 581 Type of Math Object Proof Major Section Reference Parent fundamental theorem of arithmetic Add a correction Attach a problem Ask a question Comments failure functions Permalink Submitted by akdevaraj on Thu 05 21 2015 12 24 Background In 1988 I read the book one two three infinity by George Gammow The book had a statement to the effect that no polynomial had been found such that it generates all the prime numbers and nothing but prime numbers This was true at the time Gammow wrote the book however subsequently a polynomial was constructed fulfiling the condition given above I then experimented with some polynomials and found that although one cannot generally predict the prime numbers generated by a polynomial one can predict the composite numbers generated by a polynomial Since I was originally trying to predict the primes generated by a given polynomial which may be called successes but could predict the failures composite numbers I called functions which generate failures failure functions I presented this concept at the Ramanujan Mathematical society in May 1988 Subsequently I used this tool in proving a theorem similar to the Ramanujan Nagell theorem at the AMS BENELUX meeting in 1996 Abstract definition Let f x f x f x be a function of x x x Then x g x 0 x g subscript x 0 x g x 0 is a failure function if f g x 0 is a failure in accordance with our definition of a failure Note x 0 subscript x 0 x 0 is a specific value of x x x Examples 1 Let our definition of a faiure be a composite number

Original URL path: http://www.planetmath.org/inductionproofoffundamentaltheoremofarithmetic (2016-04-25)

Open archived version from archive - Birkhoff Recurrence Theorem | planetmath.org

example you should try to find I will do this by taking a specific interval n 5 to 36 This interval is spaned by the failure functions arithmetic progressions 1 3k 2 7k 3 13k 4 7k and 5 31k 7 19k and 9 13k However the values of n skiped by the above are 5 6 8 12 14 15 17 20 21 24 27 and 33 Now all these skiped values of n are such that f n is prime So you have to find a value of n skiped by the relevant failure functions such that f n is composite Such a counter example may be very difficult to find Log in to post comments sketch proof in commonly understood terminology Permalink Submitted by akdevaraj on Sun 11 08 2015 04 45 This means we have a case of indirect primality testing f n is composite if n n 0 kf n 0 Otherwise f n is prime Note we are not testing f n directly for primality we are only testing whether is n is generated by n 0 kf n 0 if so f n is composite if not f n is prime to be continued Log in to post comments sketch proof in commonly understood terminology Permalink Submitted by akdevaraj on Sun 11 08 2015 09 25 Before proceeding let me emphasise that the values of n skiped by the failure functions i e n 0 kf n 0 are such that f n are prime and they need not be tested for primality We now set up an iteration as follows Let p 0 be the largest known prime having shape n 2 n 1 Let n 0 be the corresponding value of n i e n 0 2 n 0 1 p 0 Consider the interval n 0 n 0 p 0 I am considering this interval because f n 0 kf n 0 is congruent to 0 mod f n 0 Note the fraction of this interval not covered by the failure functions arithmetic progressions 1 3k 2 7k 3 13k 4 7k Let n 1 be the largest of these values of n not covered by any failure function f n 1 is prime which need not be tested for primality Let this prime be p 1 2nd iteration consider the interval n 1 n 1 p 1 proceed as in the 1st iteration i e mark the fraction of values of n in this interval not covered by any failure function Let n 2 be the largest value of n in this interval not covered by any failure function f n 2 is prime call this p 2 Consider the interval n 2 n 2 p 2 Mark the fraction of values of n in this interval not covered by any failure function The fraction of values of n not covered by any failure function in any interval decreases asymptotically from iteration to iteration to a rational number call

Original URL path: http://www.planetmath.org/birkhoffrecurrencetheorem (2016-04-25)

Open archived version from archive