A262228 Deficiency sequence: a(0) = 1, a(n) is the smallest prime p > a(n-1) such that the product of a(i), 1 <= i < n, is deficient (belongs to A005100).

1, 2, 5, 11, 59, 653, 84761, 2763189059, 377406001499268899, 2638619515495963542360422694651593, 135435890329895562961039215198033899386421965445591860752412324961

Offset: 0

Author

Chayim Lowen, Sep 15 2015

Keywords

• nonn

Comments

The product of the first n+1 terms is the smallest deficient multiple of the product of the first n terms.

The product of any finite number of distinct terms of this sequence is deficient.

a(n) for n > 0 is the lexicographically earliest sequence of primes P, such that the asymptotic density of the squarefree numbers (A005117) which are not divisible by any prime in P is 3/Pi^2 (A104141), i.e., half the asymptotic density of all the squarefree numbers. - Amiram Eldar, Nov 30 2020

Examples

			a(3) = 11 because A001065(2*5*7) = A001065(70) = 74 > 70, and A001065(2*5*11) = A001065(110) = 106 < 110.
From _M. F. Hasler_, Dec 14 2017: (Start)
Let Q(x) = 1/(2x/sigma(x) - 1), P(n) = Product( a(k), k<n): P(0) = 1 (empty product). Then:
Q(P(0)) = 1, a(0) = nextprime(1) = 2 = P(1).
Q(P(1)) = 3, a(1) = 5. (2*3 is perfect, P(2) = 2*5 is deficient.)
Q(P(2)) = 9, a(2) = 11. (2*5*7 is weird, P(3) = 2*5*11 is deficient.)
Q(P(3)) = 54, a(3) = 59. (P(3)*53 is weird, P(4) = 2*5*11*59 is deficient.)
Q(P(4)) = 648, a(4) = 653. (P(4)*647 is weird, P(5) = 2*5*11*59*653 is deficient.)
Q(P(5)) = 84758.4, a(5) = 84761. (P(5)*84751 is abundant and semiperfect: sum of all proper divisors except {1, 2, 11, 22, 55, 59, 590}; P(6) = 2*5*11*59*653*84761 is deficient.) (End)
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..14

Crossrefs

Cf. A001065, A005100, A005117, A104141, A151800 (nextprime).

Cf. A002975 (primitive weird numbers), A000203 (sigma), A295001 (same definition but a(0) = 4).

Programs

Mathematica

a[0]=1; a[n_] := a[n] = NextPrime[1/(2*Product[a[i],{i,1,n-1}]/Product[a[i]+1,{i,1,n-1}]-1)]; Array[a, 11, 0] (* Amiram Eldar, Jun 10 2019 *)

PARI

lista(nn) = {print1(p=1, ", "); vp = [p]; for (n=2, nn, np = nextprime(1+floor(1/(2*prod(i=2, n-1, vp[i]/(vp[i]+1))-1))); vp = concat(vp, np); print1(np, ", "););} \\ Michel Marcus, Oct 16 2015

PARI

a=List(); m=1; for(n=0, 13, listput(a, p=nextprime(1\(2/sigma(m,-1)-1)+1)); p>default(primelimit)&&addprimes(p); m*=p); a \\ M. F. Hasler, Dec 14 2017

Formula

a(n) = A151800(floor(1/(2*(Product_{i=2..n-1} a(i)/(a(i)+1))-1))), where A151800 is the "next larger prime" function.

Lim_{n->infinity} A001065(Product_{i=0..n} a(i))/(Product_{i=0..n} a(i)) = 1. [Corrected by M. F. Hasler, Dec 04 2017]

Conjecture: log(a(n)) ~ e^(an+b) where a and b are approximately 0.6 and -1.6 respectively.