OFFSET
1,1
COMMENTS
Subscript of the smallest primorial number that when divided by the (n-1)-th primorial number gives an abundant number.
Products of consecutive primes started with prime(a) up to prime(b) result in abundant squarefree numbers if b is large enough and provides perhaps the least squarefree solutions to Rivera Puzzle 329 and its generalization.
Adding a new prime p to the product increases the relative abundancy sigma(N)/N by a factor 1+1/p. This leads to a simple and fast algorithm, see the PARI code. - M. F. Hasler, Jul 30 2016
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83. (Annotated scanned copy) [Annotation on p. 82 references this A-number, but the triangle with that annotation is apparently A046900, unrelated to this entry. - Andrey Zabolotskiy, Jul 16 2022]
Carlos Rivera, Puzzle 329. Odd abundant numbers not divided by 2 or 3, The Prime Puzzles and Problems Connection.
FORMULA
EXAMPLE
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 non-deficient (being a perfect number);
n=3: a(n)=11 because prime(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.
MATHEMATICA
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 Elemer, Sep 19 2005 *)
PROG
(PARI) a=1; i=0; for(n=1, 99, while(2>a*=1+1/prime(i++), ); print1(i", "); a/=1+1/prime(n)) \\ M. F. Hasler, Jul 30 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Additional comments from Labos Elemer, Sep 19 2005
More terms from Don Reble, Nov 10 2005
Edited by N. J. A. Sloane, Dec 22 2006
STATUS
approved