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For Fibonacci numbers starting with F 1 = 0 and F 2 = 1 and with each succeeding Fibonacci number being the sum of the preceding two, one can generate a sequence of Pythagorean triples starting from (a 3, b 3, c 3) = (4, 3, 5) via
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 ...
Every prime number p divides a Fibonacci number that can be determined by the value of p modulo 5. If p is congruent to 1 or 4 modulo 5, then p divides F p−1, and if p is congruent to 2 or 3 modulo 5, then, p divides F p+1. The remaining case is that p = 5, and in this case p divides F p.
In mathematics, the Pell numbers are an infinite sequence of integers, known since ancient times, that comprise the denominators of the closest rational approximations to the square root of 2. This sequence of approximations begins 1 1 , 3 2 , 7 5 , 17 12 , and 41 29 , so the sequence of Pell numbers begins ...
1/2 + 1/4 + 1/8 + 1/16 + ⋯. First six summands drawn as portions of a square. The geometric series on the real line. In mathematics, the infinite series 1 2 + 1 4 + 1 8 + 1 16 + ··· is an elementary example of a geometric series that converges absolutely. The sum of the series is 1. In summation notation ...
That 641 is a factor of F 5 can be deduced from the equalities 641 = 2 7 × 5 + 1 and 641 = 2 4 + 5 4. It follows from the first equality that 2 7 × 5 ≡ −1 (mod 641) and therefore (raising to the fourth power) that 2 28 × 5 4 ≡ 1 (mod 641). On the other hand, the second equality implies that 5 4 ≡ −2 4 (mod 641
Adding 4 hours to 9 o'clock gives 1 o'clock, since 13 is congruent to 1 modulo 12. In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called the modulus. The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones ...
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.