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%I #34 Apr 22 2021 22:17:20
%S 3,5,9,18,31,46,67,91,122,158,194,238,284,334,392,456,522,591,668,749,
%T 835,929,1028,1133,1242,1352,1469,1594,1727,1869,2019,2163,2315,2471,
%U 2636,2802,2977,3157,3342,3534,3731,3933,4145,4358,4581,4811
%N a(n) is the least number of prime factors for any abundant number with p_n (the n-th prime) as its least factor.
%C 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]
%C 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
%C From _Jianing Song_, Apr 21 2021: (Start)
%C 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.
%C 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)
%H Amiram Eldar, <a href="/A108227/b108227.txt">Table of n, a(n) for n = 1..500</a>
%H Jianing Song, <a href="/A108227/a108227.pdf">Notes for A108227</a>
%F 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
%e 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.
%o (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
%o (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))
%o 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
%Y Cf. A000040, A005101, A006038, A001222.
%Y Cf. A107705.
%Y Cf. A007707 (~ A007684), A007708, A007741.
%Y Cf. A001276 (least number of prime factors for a (p_n)-rough abundant number, counted without multiplicity).
%K nonn
%O 1,1
%A _Hugo van der Sanden_, Jun 17 2005
%E Data corrected by _Amiram Eldar_, Aug 08 2019
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