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In particular, for a prime number p we have the explicit formula r 4 (p) = 8(p + 1). [2] Some values of r 4 (n) occur infinitely often as r 4 (n) = r 4 (2 m n) whenever n is even. The values of r 4 (n) can be arbitrarily large: indeed, r 4 (n) is infinitely often larger than . [2]
The sum of the reciprocals of the powerful numbers is close to 1.9436 . [4] The reciprocals of the factorials sum to the transcendental number e (one of two constants called "Euler's number"). The sum of the reciprocals of the square numbers (the Basel problem) is the transcendental number π 2 / 6 , or ζ(2) where ζ is the Riemann zeta ...
A box of 20 McDonald's Chicken McNuggets. One special case of the coin problem is sometimes also referred to as the McNugget numbers. The McNuggets version of the coin problem was introduced by Henri Picciotto, who placed it as a puzzle in Games Magazine in 1987, [19] and included it in his algebra textbook co-authored with Anita Wah. [20]
Combinations and permutations in the mathematical sense are described in several articles. Described together, in-depth: Twelvefold way; Explained separately in a more accessible way: Combination; Permutation; For meanings outside of mathematics, please see both words’ disambiguation pages: Combination (disambiguation) Permutation ...
4 + 1; 3 + 2; 2 + 3; 1 + 4. Compare this with the three partitions of 5 into distinct terms: 5; 4 + 1; 3 + 2. Note that the ancient Sanskrit sages discovered many years before Fibonacci that the number of compositions of any natural number n as the sum of 1's and 2's is the nth Fibonacci number!
The 2024 Billboard Music Awards are back to highlight the songs, albums and artists that dominated the charts all year round. Find out who won.
The AOL Help site is your starting point for getting support from AOL. Support may come via phone, chat, social media or help articles, depending on the question or issue you have.
The number of k-combinations for all k is the number of subsets of a set of n elements. There are several ways to see that this number is 2 n . In terms of combinations, ∑ 0 ≤ k ≤ n ( n k ) = 2 n {\textstyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}} , which is the sum of the n th row (counting from 0) of the binomial coefficients in ...