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A337814
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a(n) is the smallest primitive nondeficient number that divides n or is a multiple of n.
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1
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6, 6, 6, 20, 20, 6, 28, 88, 945, 20, 88, 6, 104, 28, 945, 272, 272, 6, 304, 20, 945, 88, 368, 6, 550, 104, 945, 28, 464, 6, 496, 1184, 3465, 272, 70, 6, 1184, 304, 4095, 20, 1312, 6, 1376, 88, 945, 368, 1504, 6, 2205, 550
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OFFSET
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1,1
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COMMENTS
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The list of primitive nondeficient numbers (A006039) starts 6, 20, 28, 70, 88, 104, 272, ..., with 945 the first odd term.
When n is a primitive nondeficient number, a(n) = n; for other values of n, either the set of divisors of n or the set of multiples of n contains a primitive nondeficient number, but not both.
If n is a nondeficient number, a(n) is a divisor: we know n is a multiple of at least one primitive nondeficient number, as it follows directly from the definition of primitive nondeficient number.
For deficient n, a(n) is a multiple. We can always find a multiple that is a primitive nondeficient number by multiplying n by the product of successive primes starting with A007918(2n-1), the first prime >= 2n-1. The smallest nondeficient number that is generated this way will be primitive, therefore an upper bound for a(n). (Reaching a nondeficient number is guaranteed because the sum of the inverses of the primes is infinite.)
More extensive explanation, due to M. F. Hasler, summarized from SeqFan list posting: (Start)
Any deficient number N has abundant multiples; to reach a primitive nondeficient number it is sufficient to choose additional prime factors in such a way that you just get abundancy >= 2, but < 2 whatever factor you omit.
An additional prime factor p increases abundancy by a factor 1 + 1/sum_{k=1..m(p)} p^k if the new multiplicity of p is m(p) >= 1.
Let x = max { sum_{k=1..m(p)} p^k : p | N } so that 1+1/x is the smallest such contribution of any prime factor in N.
Since the infinite product (over primes p) of 1+1/p diverges, a satisfactory method is multiplying N by distinct prime factors greater than x until abundancy is >= 2.
(End)
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LINKS
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FORMULA
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For m >= 1, a(6m+3) == 1 (mod 2).
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EXAMPLE
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6 is the smallest primitive nondeficient number, and is a multiple of 1, 2 and 3; so a(1) = a(2) = a(3) = 6.
For n = 4: we see 6 is not a multiple of 4, but the second smallest primitive nondeficient number, 20, is a multiple of 4; so a(4) = 20.
For n = 9: as 9 is deficient, we seek a suitable multiple. All even multiples of 9 are nondeficient, as they are multiples of nondeficient 6, but that also means they are not primitive. So we seek a qualifying odd multiple. The first odd nondeficient number is 945, which must therefore be primitive, and is also a multiple of 9. So a(9) = 945.
For n = 40: as a nondeficient number, we know 40 must have a primitive nondeficient divisor; the least such is 20, so a(40) = 20.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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