OFFSET
1,1
COMMENTS
If we replace "abundant" in the definition with "nondeficient", 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 [NOTE: As we see from n = 10, this is NOT true: see EXAMPLE! - M. F. Hasler, Jun 05 2026]
From Jianing Song, Apr 21 2021: (Start)
Let x_1 < x_2 < ... < x_k < ... be the numbers of the form p or 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 (corrected by Amiram Eldar, May 09 2026)
Jianing Song, Notes for A108227.
FORMULA
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.
For n = 10, let k = 29 * A002110(165) / A002110(9), where A002110(m) is the product of the first m primes. k has 165-9+1 = 157 prime factors (counting repetitions) with least prime factor = prime(10) = 29, and sigma(k)/k = 2.00000935..., so k is abundant. There is no abundant number with least prime factor = 29 and fewer than 157 prime factors, so a(10) = 157. - Peter Munn, May 09 2026
MATHEMATICA
formQ[p_, q_] := If[q >= p && PrimeQ[q], True, If[IntegerQ[Sqrt[4*q + 1]], Module[{r = (Floor[Sqrt[4*q + 1]] - 1)/2}, r >= p && PrimeQ[r]], False]];
a[n_] := Module[{prod = 1, sum = 0, p = Prime[n], i}, i = p; While[prod <= 2, If[formQ[p, i], prod *= (1 + 1/i); sum++]; i++]; sum];
Array[a, 50] (* Amiram Eldar, May 09 2026, after Jianing Song *)
PROG
(PARI) apply( {A108227(n)=my(p=prime(n), s=1+1/p); while(2>=s*=1+1/p=nextprime(p+1), ); n=#p=primes([prime(n), p]); p=Mat([p~, [1|p<-p]~]); for(i=1, n, while(p[i, 1]^(p[i, 2]+2)\(p[i, 1]-1) <= p[n, 1], p[i, 2]++; n--); sigma(p[1..n-1, ], -1)>2 && n--; (i<n && p[i, 2]>1)||break); vecsum(p[1..n, 2])}, [1..40]) \\ M. F. Hasler, Jun 05 2026, replacing earlier incorrect code from 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
Cf. A000040, A000203, A005101, A006038, A001222 (Omega, bigomega), A002110 (primorial numbers), A047802.
Cf. A107705 (similar for nondeficient numbers, essentially the same), A396829 (equivalent for squarefree abundant numbers).
Cf. A001276 (least number of prime factors for a (p_n)-rough abundant number, counted without multiplicity).
KEYWORD
nonn
AUTHOR
Hugo van der Sanden, Jun 17 2005
EXTENSIONS
Data aligned with 2017 PARI program output by Amiram Eldar, Aug 08 2019
Original data restored by Amiram Eldar, May 09 2026
STATUS
approved