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For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.
The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: = = (+). This equation was known ...
Genre. Mathematics, problem solving. Publication date. 1945. ISBN. 9780691164076. How to Solve It (1945) is a small volume by mathematician George Pólya, describing methods of problem solving. [1] This book has remained in print continually since 1945.
Quadratic formula. The roots of the quadratic function y = 1 2 x2 − 3x + 5 2 are the places where the graph intersects the x -axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
Knuth's up-arrow notation. In mathematics, Knuth's up-arrow notation is a method of notation for very large integers, introduced by Donald Knuth in 1976. [1] In his 1947 paper, [2] R. L. Goodstein introduced the specific sequence of operations that are now called hyperoperations. Goodstein also suggested the Greek names tetration, pentation ...
For example, let a denote a multiplicative generator of the group of units of F 4, the Galois field of order four (thus a and a + 1 are roots of x 2 + x + 1 over F 4. Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits ...
Given a polygon P with n + 2 sides and a triangulation, mark one of its sides as the base, and also orient one of its 2n + 1 total edges. There are (4n + 2)C n such marked triangulations for a given base. Given a polygon Q with n + 3 sides and a (different) triangulation, again mark one of its sides as the base. Mark one of the sides other than ...
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