OFFSET
1,1
COMMENTS
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<=i<r} prime(n+i) for some r, which turns out to be r = a(n). - M. 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)
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..500
Jianing Song, Notes for A108227
FORMULA
a(n) = A007684(n)-n+1, for n>1. A007741(n) = Product_{0<=i<a(n)} prime(n+i). - M. F. Hasler, Jun 15 2017
EXAMPLE
a(2) = 5 since 945 = 3^3*5*7 is an abundant number with p_2 = 3 as its smallest prime factor, and no such number exists with fewer than 5 prime factors.
PROG
(PARI) A108227(n, s=1+1/prime(n))=for(a=1, 9e9, if(2<s*=1+1/prime(n+a), return(a+1))) \\ M. F. Hasler, Jun 15 2017
(PARI) isform(k, q) = my(p=prime(k)); if(isprime(q) && (q>=p), 1, if(issquare(4*q+1), my(r=(sqrtint(4*q+1)-1)/2); isprime(r) && (r>=p), 0))
a(n) = my(Prod=1, Sum=0); for(i=prime(n), oo, if(isform(n, i), Prod *= (1+1/i); Sum++); if(Prod>2, return(Sum))) \\ Jianing Song, Apr 21 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Hugo van der Sanden, Jun 17 2005
EXTENSIONS
Data corrected by Amiram Eldar, Aug 08 2019
STATUS
approved