cp's OEIS Frontend

A number n is nondeficient (A023196) iff it is abundant or perfect, that is iff A001065(n) is >= n. Since any multiple of a nondeficient number is itself nondeficient, we call a nondeficient number primitive iff all its proper divisors are deficient. - Jeppe Stig Nielsen, Nov 23 2003

Numbers whose proper multiples are all abundant, and whose proper divisors are all deficient. - Peter Munn, Sep 08 2020

As a set, shares with the sets of k-almost primes this property: no member divides another member and each positive integer not in the set is either a divisor of 1 or more members of the set or a multiple of 1 or more members of the set, but not both. See A337814 for proof etc. - Peter Munn, Apr 13 2022

See A006039 for a definition and list of primitive nondeficient numbers.

The first prime being 2, the prime(1)-smooth numbers are the powers of 2, which are all deficient. So a(1) is undefined, and the sequence offset is 2.

Omitting the initial "6" gives us the largest prime(n)-smooth primitive abundant numbers (based on their A071395 definition). Using the variant definition of primitive abundant from A091191, the equivalent sequence starts 18, 30, 2205, 12705, 117234117, ... .

If m is a prime(n)-smooth primitive nondeficient number, the odd part of m divides a member of one of the first (n - 1) finite sets described in the Dickson reference and the even part of m is less than 2^A035100(n). This provides an upper bound for such numbers, meaning there is a largest prime(n)-smooth primitive nondeficient number for all n >= 2.

This triangle is the analog of A188439 for A001222 ("bigomega", total number of prime factors) instead of A001221 ("omega", distinct prime divisors). It starts with row 5, since there is no odd primitive abundant number, N in A006038, with less than A001222(N) = 5 prime factors (counted with multiplicity).

Sequence A287728 gives the row lengths: Row 5 has 121 terms (945, 1575, 2205, 3465, 4095, ..., 430815, 437745, 442365). This mostly equals the initial terms of A006038, except for those with indices {12, 39, 40, 45, 48, 54, ..., 87}. These are in turn mostly (except for the 17th and 18th term) those of the subsequent row 6 which has 15772 terms, (7425, 28215, 29835, 33345, 34155, ..., 13443355695, 13446051465, 13455037365).

Sequences A275449 and A287581 give the smallest and largest* element of each row (*assuming that the largest term in the row is squarefree). Accordingly, row 7 starts with A275449(7) = 81081, and ends with A287581(7) = 1725553747427327895.

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)