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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! Note that these are not general compositions as defined above because the numbers are restricted to 1's and 2's only. 1=1 (1) 2=1+1=2 (2) 3=1+1+1=1+2=2+1 (3)
A number n that has more divisors than any x < n is a highly composite number (though the first two such numbers are 1 and 2). Composite numbers have also been called "rectangular numbers", but that name can also refer to the pronic numbers, numbers that are the product of two consecutive integers. Yet another way to classify composite numbers ...
A related concept is that of a largely composite number, a positive integer that has at least as many divisors as all smaller positive integers. The name can be somewhat misleading, as the first two highly composite numbers (1 and 2) are not actually composite numbers; however, all further terms are. Ramanujan wrote a paper on highly composite ...
Every positive integer greater than 1 is either the product of two or more integer factors greater than 1, in which case it is a composite number, or it is not, in which case it is a prime number. For example, 15 is a composite number because 15 = 3 · 5 , but 7 is a prime number because it cannot be decomposed in this way.
As the number of bits composing a string increases, the number of possible 0 and 1 combinations increases exponentially. A single bit allows only two value-combinations, two bits combined can make four separate values, three bits for eight, and so on, increasing with the formula 2 n. The amount of possible combinations doubles with each binary ...
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These 2 Nutrients Could Help You Lose Weight Tanja Ivanova - Getty Images. There’s a lot of information out there about how to lose weight—and some tidbits are more reliable than others.
The rightmost digit is 2−1 = 1. The second digit from the right would become −1, so add 4 to give 3 and then carry +1 two places to the left. The third digit from the right is 1−0 = 1. Then the leftmost digit is 1−1 plus 1 from the carry, giving 1. This gives a final answer of .
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