1,1

Subscript of the smallest primorial number which when divided by the (n-1)-th primorial number gives an abundant number.

Products of consecutive primes started with p(a) up to p(b) result in abundant square-free numbers if b is large enough and provides perhaps the least square-fre solutions to Rivera Puzzle 329 and its generalization.

Walter Nissen, Home Page (listed in lieu of email address)

C. Rivera, Puzzle 329. Odd abundant numbers not divided by 2 or 3.

a(n) is the solution to Min{x : Floor[sigma(p#(x)/p#(n-1))/(p#(x)/p#(n-1))=2]}, where p#(w) is the w-th primorial number, the product of first w prime numbers. For a>b, the p#(a)/p#(b)=A000210(a)/A000210(b) quotients are integer numbers as follows: p(b+1)*p(b+2)...p(a), where p(j) is the j-th prime number.

n=1: a(1)=2 means that primorial(2)=6 divided by primorial(1-1)=1 gives the quotient 6/1=6 which is just abundant (being a perfect number);

n=3: a(n)=11 because p(3)=5 primorial(11)=2.3.5....29.31,primorial(3-1)=2.3=6.

p#(11)/p#(2)=3.5.7.11.13.17.19.23.29.31= 33426748355 = q and sigma(q)/q = 2.00097 >2 so q is an abundant number.Also p#(10)/p#(3-1) is not yet abundant.

spr[x_, y_] :=Apply[Times, Table[(Prime[w]+1)/(Prime[w]), {w, x, y}]] Table[Min[Flatten[Position[Table[Floor[spr[n, w]], {w, 1, 1000}], 2]]], {n, 1, 20}] (Labos)

Cf. A005100, A007686, A007702, A007707 (an essentially identical sequence).

Cf. A000210, A064001, A112640, A005101, A005231.

Sequence in context: A085573 A081691 A085571 this_sequence A135348 A083322 A073939

Adjacent sequences: A007681 A007682 A007683 this_sequence A007685 A007686 A007687

nonn

Walter Nissen

Additional comments from Labos E. (labos(AT)ana.sote.hu), Sep 19 2005

More terms from Don Reble (djr(AT)nk.ca), Nov 10 2005

Edited by N. J. A. Sloane (njas(AT)research.att.com), Dec 22 2006