Pascal`s triangle and the binomial theorem
... 4. The binomial theorem If we wanted to expand a binomial expression with a large power, e.g. (1 + x)32 , use of Pascal’s triangle would not be recommended because of the need to generate a large number of rows of the triangle. An alternative method is to use the binomial theorem. The theorem enabl ...
... 4. The binomial theorem If we wanted to expand a binomial expression with a large power, e.g. (1 + x)32 , use of Pascal’s triangle would not be recommended because of the need to generate a large number of rows of the triangle. An alternative method is to use the binomial theorem. The theorem enabl ...
DECISION PROBLEMS OF FINITE AUTOMATA DESIGN
... 1. Motivation. Many variants of the notion of automaton have appeared in the literature. We find it convenient here to adopt the notion of E. F. Moore [7]. Inasmuch as Rabin-Scott [9] adopt this notion, too, it is convenient to refer to [9] for various results presumed here. In particular, Kleene's ...
... 1. Motivation. Many variants of the notion of automaton have appeared in the literature. We find it convenient here to adopt the notion of E. F. Moore [7]. Inasmuch as Rabin-Scott [9] adopt this notion, too, it is convenient to refer to [9] for various results presumed here. In particular, Kleene's ...
39(3)
... with new ideas that develop enthusiasm for number sequences or the exploration of number facts. Illustrations and tables should be wisely used to clarify the ideas of the manuscript. Unanswered questions are encouraged, and a complete list of references is absolutely necessary. ...
... with new ideas that develop enthusiasm for number sequences or the exploration of number facts. Illustrations and tables should be wisely used to clarify the ideas of the manuscript. Unanswered questions are encouraged, and a complete list of references is absolutely necessary. ...
Full text
... L e m m a 2 . 1 : If j is represented as the sum of distinct powers of 2 in descending order, j = 2r + 2s + j- 2 W , r > 5 > w, then the j t h even-Zeck integer N = N(j) is given by Zeck N(j) = F2(r+i) + -^2(5+1) H 1" -^(w+i)- I n short, replace the binary representation of j in the powers 2p,p = 0 ...
... L e m m a 2 . 1 : If j is represented as the sum of distinct powers of 2 in descending order, j = 2r + 2s + j- 2 W , r > 5 > w, then the j t h even-Zeck integer N = N(j) is given by Zeck N(j) = F2(r+i) + -^2(5+1) H 1" -^(w+i)- I n short, replace the binary representation of j in the powers 2p,p = 0 ...
What is . . . tetration?
... Remark 1.12. The proof appears to still hold if “rational number” in the statement of the corollary is replaced with “algebraic.” According to [AT] the question for n a, n cases n = 2 and n = 3. ...
... Remark 1.12. The proof appears to still hold if “rational number” in the statement of the corollary is replaced with “algebraic.” According to [AT] the question for n a, n cases n = 2 and n = 3. ...
The Fibonacci Numbers And An Unexpected Calculation.
... By design, ConfuseL can’t be on the list L! ConfuseL differs from the kth element on the list L in the kth position. This contradicts the assumption that the list L is complete; i.e., that the map f: to R[0,1] is surjective. ...
... By design, ConfuseL can’t be on the list L! ConfuseL differs from the kth element on the list L in the kth position. This contradicts the assumption that the list L is complete; i.e., that the map f: to R[0,1] is surjective. ...
Chebyshev`s conjecture and the prime number race
... π(x, 8, 1) 6 maxa∈{3,5,7} π(x, 8, a) for x 6 106 . It is generally believed that the answers to both questions are yes. If r = 2 this is just the first problem (infinity of sign changes). 8. “Distribution problems”. Investigate the distribution of ∆(x, k, l1 , l2 ), e.g. study S(x; z) = {1 6 y 6 x : ...
... π(x, 8, 1) 6 maxa∈{3,5,7} π(x, 8, a) for x 6 106 . It is generally believed that the answers to both questions are yes. If r = 2 this is just the first problem (infinity of sign changes). 8. “Distribution problems”. Investigate the distribution of ∆(x, k, l1 , l2 ), e.g. study S(x; z) = {1 6 y 6 x : ...
Full text
... Let an denote the number of compositions of n subject to some system of constraints C. If the constraint is using only odd parts, then an = Fn (the nth Fibonacci number). Thus, 5 = 3 + 1 + 1 = 1 + 3 + 1 = 1 + 1 + 3 = 1 + 1 + 1 + 1 + 1 has 5 = F5 compositions. If the constraint is using only parts ≥ ...
... Let an denote the number of compositions of n subject to some system of constraints C. If the constraint is using only odd parts, then an = Fn (the nth Fibonacci number). Thus, 5 = 3 + 1 + 1 = 1 + 3 + 1 = 1 + 1 + 3 = 1 + 1 + 1 + 1 + 1 has 5 = F5 compositions. If the constraint is using only parts ≥ ...
Prime Numbers in digits of `e`
... digit prime number in the digits of e. First a Taylor series will calculate a value of e through 500 iterations of the series. Then using the Lucas Primality Test each ten consecutive digits will be sampled to test for primality. It is hard to tell if the arbitrary settings of 500 iterations and 40 ...
... digit prime number in the digits of e. First a Taylor series will calculate a value of e through 500 iterations of the series. Then using the Lucas Primality Test each ten consecutive digits will be sampled to test for primality. It is hard to tell if the arbitrary settings of 500 iterations and 40 ...
Lesson 8-1 Geometric Mean with answers.notebook
... For two positive numbers a and b, the geometric mean is the positive number x where the proportion a : x = x : b is a x true. This proportion can be written as fractions = x b or with cross products x2 = ab or x = √ab ...
... For two positive numbers a and b, the geometric mean is the positive number x where the proportion a : x = x : b is a x true. This proportion can be written as fractions = x b or with cross products x2 = ab or x = √ab ...
Polygonal Numbers and Finite Calculus
... Figure 2 gives numerical and geometric representations of triangular and pentagonal, both also polygonal. As with square numbers, triangular and pentagonal numbers possess the first characteristic, the capacity to be represented by arrangements of equally spaced points that form regular 3- and 5-gon ...
... Figure 2 gives numerical and geometric representations of triangular and pentagonal, both also polygonal. As with square numbers, triangular and pentagonal numbers possess the first characteristic, the capacity to be represented by arrangements of equally spaced points that form regular 3- and 5-gon ...
Chapter 12 - Princeton University Press
... subsection for general k. According to [HW], note on Chapter XXIII, the theorem for k = 1 was discovered independently by Bohl, Sierpiński and Weyl at about the same time. We follow Weyl’s proof which, according to [HW], is undoubtedly “the best proof” of the theorem. There are other proofs, howeve ...
... subsection for general k. According to [HW], note on Chapter XXIII, the theorem for k = 1 was discovered independently by Bohl, Sierpiński and Weyl at about the same time. We follow Weyl’s proof which, according to [HW], is undoubtedly “the best proof” of the theorem. There are other proofs, howeve ...
Math 319 Problem Set #6 – Solution 5 April 2002
... Factorizations beginning with 3 may be ignored, because 3 does not occur as ϕ of any prime power. Factorizations beginning with 4: The only such factorization with non-decreasing factors is 4 × 12. This can arise in three ways: 4 × 12 = ϕ(5)ϕ(13) = ϕ(2)ϕ(5)ϕ(13) = ϕ(8)ϕ(13) Factorizations beginning ...
... Factorizations beginning with 3 may be ignored, because 3 does not occur as ϕ of any prime power. Factorizations beginning with 4: The only such factorization with non-decreasing factors is 4 × 12. This can arise in three ways: 4 × 12 = ϕ(5)ϕ(13) = ϕ(2)ϕ(5)ϕ(13) = ϕ(8)ϕ(13) Factorizations beginning ...
Absolute Primes
... N0 = c1 . . . cn−4 7931 = L · 105 + 7931, where the notation a1 . . . an is used to denote the number a1 10n−1 +a2 10n−2 + . . . + an−1 10 + an , which decimal representation consists of digits a1 , . . . , an . The integers K0 = 7931, K1 = 1793, K2 = 9137, K3 = 7913, K4 = 7193, K5 = 1937, K6 = 7139 ...
... N0 = c1 . . . cn−4 7931 = L · 105 + 7931, where the notation a1 . . . an is used to denote the number a1 10n−1 +a2 10n−2 + . . . + an−1 10 + an , which decimal representation consists of digits a1 , . . . , an . The integers K0 = 7931, K1 = 1793, K2 = 9137, K3 = 7913, K4 = 7193, K5 = 1937, K6 = 7139 ...
A remark on the extreme value theory for continued fractions
... Proof. Suppose the liminf is positive, otherwise the result is obvious. We may assume that each interval in En−1 contains exactly mn intervals of En since we can remove some excess intervals to get smaller sets En0 and E 0 . Thus the gaps of lengths between different intervals in En are not changed ...
... Proof. Suppose the liminf is positive, otherwise the result is obvious. We may assume that each interval in En−1 contains exactly mn intervals of En since we can remove some excess intervals to get smaller sets En0 and E 0 . Thus the gaps of lengths between different intervals in En are not changed ...
(pdf)
... partitions of n with parts at most k (in value). We now form a bijection between the two sets, associating with each member of S the partition represented by the conjugate of its Ferrers diagram, which we claim is a member of T . Since a partition with no more than k parts will be represented by a d ...
... partitions of n with parts at most k (in value). We now form a bijection between the two sets, associating with each member of S the partition represented by the conjugate of its Ferrers diagram, which we claim is a member of T . Since a partition with no more than k parts will be represented by a d ...
Absolutely Abnormal Numbers - Mathematical Association of America
... bers that are simply normalto no base whatsoever.As we saw in the proof of our Theorem,the numbera does meet this strongercriterionof abnormality.On the other hand,while all rationalnumbersare absolutelyabnormal,many of them are in fact simplynormalto variousbases.Forexample,1/3 is simplynormalto th ...
... bers that are simply normalto no base whatsoever.As we saw in the proof of our Theorem,the numbera does meet this strongercriterionof abnormality.On the other hand,while all rationalnumbersare absolutelyabnormal,many of them are in fact simplynormalto variousbases.Forexample,1/3 is simplynormalto th ...