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Observe that the condition N/k > (k - 1)/2 is equivalent to 2N/k > (k - 1) and, since k is odd while 2N/k is even, it is equivalent to 2N/k > k, i.e., k √ 2N, which leads to the following Theorem But let's reformulate what has been proved. If m is even, then m = 2N/k, where k = 2a + m + 1, giving a decomposition of kind (2) of length 2N/(2a + m + 1) = m. If m is odd then the decomposition is of kind (1) of length m. And, since this is a decomposition of N, only of them is an odd factor of N. Since this decomposition adds up to m(2a + m + 1)/2, and the factors m and (2a + m + 1) are of different parities, only one of them is odd. , a + m) be a decomposition of N for some a and m. We now need to show that every decomposition of N is in one of the forms, (1) or (2). Indeed, the number of terms that have been dropped equals 2 + 1 = k - 2N/k, while the length of the decomposition in (1) is k. Since the first term in (1) N/k - (k - 1)/2 will have to be canceled with (k - 1)/2 - N/k, the first positive term left over will be (k - 1)/2 - N/k + 1, starting the decomposition In this case, we may drop 0, if one is present, and cancel negative terms on the left of the sum with their positive counterparts across 0. In this case, we already have a valid decomposition. We have to consider two cases depending on whether the first term in the so obtained decomposition is negative or not. Adding to each N/k produces k numbers (1) Since the powers of 2 (and only them) have no odd factors, this will prove the theorem. As an example, number 15 has three odd decompositions (15), (4, 5, 6), and (1, 2, 3, 4, 5), and one even, (7, 8).įor the proof, we shall construct a 1-1 correspondence between the odd factors of number N and its decompositions. A decomposition of an odd (even) length is called odd ( even). A decomposition that consists of a single term is trivial. The number of terms in the decomposition is its length. Theoremįor convenience, we shall call a representation of a number into the sum of consecutive integers decomposition. Still, clearly the powers of 2 stand out. The constructs for the even numbers are not so easily classified. The 2-term sums for odd numbers are pretty obvious: 2n + 1 = n + (n + 1). To get a taste for the problem, consider a few examples. Does not this remind one of the Epimenides' paradox? May we not declare the absence of a property unproperty? Which is of cause a property in its own right. So the powers of 2 do possess a property which is the absence of the property in question. The absence of this particular property is a characteristic feature of the powers of 2. The elements of the former possess the property of being the sum of two or more consecutive integers the elements in the complement do not possess that property. Those that are form a set and those that are not form its complement. A number may or may not be the sum of two or more consecutive integers. For, it is clear that a set and its complement carry exactly the same amount of information. The author sets out to answer the following question:Īnd the answer is: all natural numbers, but the powers of 2. There are of course additional properties.Ī curious result has been published 2007 that deals with a property common to all natural numbers, except for the powers of 2. Also, only for the sets whose size is a power of two, the multisets formed by the sum of pairs of elements may be equal. For example, the integer iterations on a circle that generate Ducci's sequence always settle to all zeros in a finite number of steps only if the starting sequence counts the number of terms which is a power of 2.
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Using std::pow using std::ostream_iterator Ĭout arr = In the following example, we raise each element of an int vector to the arbitrary power of 3. pow can’t be used to calculate the root of a negative number, and instead, we should utilize std::sqrt or std::cbrt. Note that this function has multiple exceptions and special cases that need to be handled by the programmer or implemented using the separate function provided in the C++ library header. The std::pow function can be used to compute the product of a given base number raised to the power of n, where n can be an integral or floating-point value. Use the std::pow Function to Raise a Number to the Power in C++
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#Number is apower of 2 how to
This article will demonstrate multiple methods of how to raise a number to the power in C++.