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The smallest odd integer with abundancy index exceeding 3 is 1018976683725 = 3 3 × 5 2 × 7 2 × 11 × 13 × 17 × 19 × 23 × 29. [8] If p = (p 1, ..., p n) is a list of primes, then p is termed abundant if some integer composed only of primes in p is abundant. A necessary and sufficient condition for this is that the product of p i /(p i − ...
Let Δ be a negative integer with Δ = −dn, where d is a multiplier and Δ is the negative discriminant of some quadratic form. Take the t first primes p 1 = 2, p 2 = 3, p 3 = 5, ..., p t, for some t ∈ N. Let f q be a random prime form of G Δ with ( Δ / q ) = 1. Find a generating set X of G Δ.
In number theory, an n-smooth (or n-friable) number is an integer whose prime factors are all less than or equal to n. [1] [2] For example, a 7-smooth number is a number in which every prime factor is at most 7. Therefore, 49 = 7 2 and 15750 = 2 × 3 2 × 5 3 × 7 are both 7-smooth, while 11 and 702 = 2 × 3 3 × 13 are not 7-smooth.
The next odd divisor to be tested is 7. One has 77 = 7 · 11, and thus n = 2 · 3 2 · 7 · 11. This shows that 7 is prime (easy to test directly). Continue with 11, and 7 as a first divisor candidate. As 7 2 > 11, one has finished. Thus 11 is prime, and the prime factorization is; 1386 = 2 · 3 2 · 7 · 11.
The entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right complex half plane with absolute value of the real part larger than or equal to the absolute value of the imaginary part.
An odd number does not have the prime factor 2. The first: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23 (sequence A005408 in the OEIS). All integers are either even or odd. A square has even multiplicity for all prime factors (it is of the form a 2 for some a). The first: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 (sequence A000290 in the OEIS).
If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as 1×2, 2×1, (−1)×(−2), or (−2)×(−1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values p(0) = 1, 2, −1, or −2. and likewise for p(1).
The number of domino tilings of a 4×4 checkerboard is 36. [10] Since it is possible to find sequences of 36 consecutive integers such that each inner member shares a factor with either the first or the last member, 36 is an Erdős–Woods number. [11] The sum of the integers from 1 to 36 is 666 (see number of the beast). 36 is also a ...