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Search: id:A090885
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| A090885 |
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Sum of the squares of the exponents in the prime factorization of n. |
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+0
1
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| 0, 1, 1, 4, 1, 2, 1, 9, 4, 2, 1, 5, 1, 2, 2, 16, 1, 5, 1, 5, 2, 2, 1, 10, 4, 2, 9, 5, 1, 3, 1, 25, 2, 2, 2, 8, 1, 2, 2, 10, 1, 3, 1, 5, 5, 2, 1, 17, 4, 5, 2, 5, 1, 10, 2, 10, 2, 2, 1, 6, 1, 2, 5, 36, 2, 3, 1, 5, 2, 3, 1, 13, 1, 2, 5, 5, 2, 3, 1, 17, 16, 2, 1, 6, 2, 2, 2, 10, 1, 6, 2, 5, 2, 2, 2, 26, 1, 5, 5, 8
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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Contribution from Daniel Forgues (squid(AT)zensearch.com), Mar 30 2009: (Start)
Euclidean norm (square of the length as measured from the origin 0 which represents the number 1) of the exponents vector of n.
If we consider n as represented as an exponents vector in an infinite dimensional discrete vector space (infinite dimensional lattice) where each dimension corresponds to a prime {p1, p2, p3, p4, p5, p6, ...} = {2, 3, 5, 7, 11, 13, ...} then the product of n1 with n2 corresponds to vector addition of the exponents vectors of n1 and n2.
If 2 numbers n1 and n2 are coprime then the length of the exponents vector of the product n1*n2 is the Pythagorean sum of the lengths of the exponents vectors of n1 and n2.
For the product of 2 arbitrary numbers n1 and n2 we have the triangle inequality applying to the lengths of the exponents vectors of n1, n2, n1*n2. E.g. 107653 = 7^2 * 13^3 is represented as (0, 0, 0, 2, 0, 3, 0, 0, 0, ...) as an exponents vector in an infinite dimensional space associated with the primes.
If all the coordinates of the exponents vector are positive we have the representation of an integer. If some components are negative then we have the representation of a rational number. The origin 0 corresponds to the number 1. There is no representation for 0 as an exponents vector.
If 2 numbers are coprime then their exponents vectors are orthogonal. If the exponents vectors of 2 numbers n1 and n2 are parallel then we have n1^a = n2^b for some non-zero integers a and b. (End)
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LINKS
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Daniel Forgues, Table of n, a(n) for n=1..100000
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FORMULA
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Additive with a(p^e) = e^2.
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CROSSREFS
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Sequence in context: A069098 A126241 A019777 this_sequence A008476 A112621 A081448
Adjacent sequences: A090882 A090883 A090884 this_sequence A090886 A090887 A090888
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KEYWORD
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easy,nonn
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AUTHOR
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Sam Alexander (amnalexander(AT)yahoo.com), Dec 12 2003
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EXTENSIONS
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More terms from Ray Chandler (rayjchandler(AT)sbcglobal.net), Dec 20 2003
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