OFFSET
0,2
COMMENTS
Formula: a(n) = 1 + [..[[[[n*2/1]3/2]4/3]5/4]...(k+1)/k]...] where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 6); thus the product will eventually reach a maximum value of a(n).
FORMULA
a(n) = 1 + 6*A073363(n).
EXAMPLE
Sieve starts with the natural numbers:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,...
Step 1: keep 1 term, remove the next 5, repeat; giving
1,7,13,19,25,31,37,43,49,55,61,67,73,79,...
Step 2: keep 2 terms, remove the next 5, repeat; giving
1,7,43,49,85,91,127,133,169,175,211,217,...
Step 3: keep 3 terms, remove the next 5, repeat; giving
1,7,43,169,175,211,337,343,379,505,511,547,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2) = 1+[..[(2)*2/1]*3/2]*4/3]*5/4]*7/6]*8/7]*9/8]*10/9]*
12/11]*13/12]*14/13]*15/14]*17/16]*18/17]*19/18]*20/19]*
22/21]*23/22]*24/23]*25/24]*27/26]*28/27]*29/28]*30/29] =43.
PROG
(PARI) {a(n)=local(A=n, B=0, k=0); until(A==B, k=k+1; if(k%6==0, k=k+1); B=A; A=floor(A*(k+1)/k)); 1+A}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 14 2005
STATUS
approved