A337814 a(n) is the smallest primitive nondeficient number that divides n or is a multiple of n.

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

Offset: 1

Author

Peter Munn, Sep 23 2020

Keywords

• nonn

Comments

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)

Examples

			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.
		

Links

• M. F. Hasler and Peter Munn, A property of k-almost primes, but also A006039?, SeqFan list, Oct 09 2020.
• Wikipedia, Divergence of the sum of the reciprocals of the primes.

Crossrefs

Sequences with related definitions: A064162, A254572.

Range of values: A006039.

See A000203, A005100 and A023196 for definitions of deficient and nondeficient numbers.

Cf. A000040, A007918, A308710, A338133.

Formula

For m >= 2, a(A000040(m)) = A338133(m, 1).

For m >= 1, a(6m+3) == 1 (mod 2).