A108227 a(n) is the least number of prime factors for any abundant number with p_n (the n-th prime) as its least factor.

3, 5, 9, 18, 31, 46, 67, 91, 122, 158, 194, 238, 284, 334, 392, 456, 522, 591, 668, 749, 835, 929, 1028, 1133, 1242, 1352, 1469, 1594, 1727, 1869, 2019, 2163, 2315, 2471, 2636, 2802, 2977, 3157, 3342, 3534, 3731, 3933, 4145, 4358, 4581, 4811

Offset: 1

Author

Hugo van der Sanden, Jun 17 2005

Keywords

• nonn

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<=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)

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..500
• Jianing Song, Notes for A108227

Crossrefs

Cf. A000040, A005101, A006038, A001222.

Cf. A107705.

Cf. A007707 (~ A007684), A007708, A007741.

Cf. A001276 (least number of prime factors for a (p_n)-rough abundant number, counted without multiplicity).

Programs

PARI

A108227(n, s=1+1/prime(n))=for(a=1, 9e9, if(2M. 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

Formula

a(n) = A007684(n)-n+1, for n>1. A007741(n) = Product_{0<=iM. F. Hasler, Jun 15 2017

Extensions

Data corrected by Amiram Eldar, Aug 08 2019