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For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2. If the initial term of an arithmetic progression is a 1 {\displaystyle a_{1}} and the common difference of successive members is d {\displaystyle d} , then the n {\displaystyle n} -th term of the sequence ( a n {\displaystyle a_{n ...
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
"subtract if possible, otherwise add": a(0) = 0; for n > 0, a(n) = a(n − 1) − n if that number is positive and not already in the sequence, otherwise a(n) = a(n − 1) + n, whether or not that number is already in the sequence.
(sequence A019279 in the OEIS). To illustrate: it can be seen that 16 is a superperfect number as σ(16) = 1 + 2 + 4 + 8 + 16 = 31, and σ(31) = 1 + 31 = 32, thus σ(σ(16)) = 32 = 2 × 16. If n is an even superperfect number, then n must be a power of 2, 2 k, such that 2 k+1 − 1 is a Mersenne prime. [1] [2]
An integer sequence is computable if there exists an algorithm that, given n, calculates a n, for all n > 0. The set of computable integer sequences is countable.The set of all integer sequences is uncountable (with cardinality equal to that of the continuum), and so not all integer sequences are computable.
For example, the sequence of powers of two (1, 2, 4, 8, ...), the basis of the binary numeral system, is a complete sequence; given any natural number, we can choose the values corresponding to the 1 bits in its binary representation and sum them to obtain that number (e.g. 37 = 100101 2 = 1 + 4 + 32). This sequence is minimal, since no value ...
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A bijection with the sums to n is to replace 1 with 0 and 2 with 11. The number of binary strings of length n without an even number of consecutive 0 s or 1 s is 2F n. For example, out of the 16 binary strings of length 4, there are 2F 4 = 6 without an even number of consecutive 0 s or 1 s—they are 0001, 0111, 0101, 1000, 1010, 1110. There is ...