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  2. Formulas for generating Pythagorean triples - Wikipedia

    en.wikipedia.org/wiki/Formulas_for_generating...

    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

  3. 1 + 2 + 3 + 4 + ⋯ - Wikipedia

    en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E...

    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 ...

  4. Fibonacci sequence - Wikipedia

    en.wikipedia.org/wiki/Fibonacci_sequence

    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.

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  6. 1/2 + 1/4 + 1/8 + 1/16 + ⋯ - Wikipedia

    en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/...

    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 ...

  7. Modular arithmetic - Wikipedia

    en.wikipedia.org/wiki/Modular_arithmetic

    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 ...

  8. Fermat number - Wikipedia

    en.wikipedia.org/wiki/Fermat_number

    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 41 (mod 641). On the other hand, the second equality implies that 5 4 ≡ −2 4 (mod 641

  9. 1 − 2 + 3 − 4 + ⋯ - Wikipedia

    en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88...

    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.