cp's OEIS Frontend

Essentially, i.e., except for a(1), identical to A007702. All terms are primitive abundant numbers (A091191) and thus, except for the first term, odd primitive abundant (A006038). The next term is too large to be displayed here, see A007707 (and formula) for many more terms, using a more compact encoding. - M. F. Hasler, Apr 30 2017

Differs from A007686 only for n=1. - Michel Marcus, Mar 10 2013

If we replace "abundant" in the definition with "non-deficient", we get the same sequence with an initial 2 instead of 3, barring an astronomically unlikely coincidence with some as-yet-undiscovered odd perfect number. [This is sequence A107705. - M. F. Hasler, Jun 14 2017]

It appears that all terms >= 5 correspond to the odd primitive abundant numbers (A006038) which are products of consecutive primes (cf. A285993), i.e., of the form N = Product_{0<=iM. F. Hasler, May 08 2017

From Jianing Song, Apr 21 2021: (Start)

Let x_1 < x_2 < ... < x_k < ... be the numbers of the form p of p^2 + p, where p is a prime >= prime(n). Then a(n) is the smallest N such that Product_{i=1..N} (1 + 1/x_i) > 2. See my link below for a proof.

For example, for n = 3, we have {x_1, x_2, ..., x_k, ...} = {5, 7, 11, 13, 17, 19, 23, 29, 5^2 + 5, ...}, we have Product_{i=1..8} (1 + 1/x_i) < 2 and Product_{i=1..9} (1 + 1/x_i) > 2, so a(3) = 9. (End)

Subscript of the smallest primorial number that when divided by the (n-1)-th primorial number gives an abundant number.

Products of consecutive primes started with prime(a) up to prime(b) result in abundant squarefree numbers if b is large enough and provides perhaps the least squarefree solutions to Rivera Puzzle 329 and its generalization.

Adding a new prime p to the product increases the relative abundancy sigma(N)/N by a factor 1+1/p. This leads to a simple and fast algorithm, see the PARI code. - M. F. Hasler, Jul 30 2016

Barring unforeseen odd perfect numbers (which it has been proved must have at least 29 prime factors if they exist at all), if we replace "non-deficient" in the description with "abundant", the value of a(1) becomes 3 and all other values stay the same.

The above mentioned sequence is A108227, see there for a comment on the relation of this sequence to that of primitive abundant numbers (A006038) which are products of consecutive primes, i.e., of the form N = Product_{0<=iA007702. - M. F. Hasler, Jun 15 2017

The smallest term is a(5) = 3*5*7*11*13, there is no odd abundant number (A005231) equal to the product of less than 5 consecutive primes.

The smallest odd abundant number (A005231) equal to the product of n consecutive primes is equal (when it exists, i.e., for n >= 5) to the least odd number with n (distinct) prime divisors, equal to the product of the first n odd primes = A070826(n+1) = A002110(n+1)/2.

See A188342 = (945, 3465, 15015, 692835, 22309287, ...) for the least odd primitive abundant number (A006038) with n distinct prime factors, and A275449 for the least odd primitive abundant number with n prime factors counted with multiplicity.

The terms are in general not primitive abundant numbers (A091191), in particular this cannot be the case when a(n) is a multiple of a(n-1), as is the case for most of the terms, for which a(n) = a(n-1)*A117366(a(n-1)). In the other event, spf(a(n)) = nextprime(spf(a(n-1))), and a(n) is in A007741(2,3,4...). These are exactly the primitive terms in this sequence.

The smallest term is a(5), there is no odd abundant number (A005231) equal to the product of less than 5 consecutive primes.

The corresponding abundant numbers are A285993(n) = prime(k-n+1)*...*prime(k), with prime(k) = a(n).