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 5); thus the product will eventually reach a maximum value of a(n).
FORMULA
a(n) = 1 + 5*A073362(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 4, repeat; giving
1,6,11,16,21,26,31,36,41,46,51,56,61,66,...
Step 2: keep 2 terms, remove the next 4, repeat; giving
1,6,31,36,61,66,91,96,121,126,151,156,...
Step 3: keep 3 terms, remove the next 4, repeat; giving
1,6,31,96,121,126,211,216,241,306,331,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2) = 1 + [..[(2)*2/1]*3/2]*4/3]*6/5]*7/6]*8/7]*10/9]*
11/10]*12/11]*14/13]*15/14]*16/15]*18/17]*19/18]*20/19] = 21.
PROG
(PARI) {a(n)=local(A=n, B=0, k=0); until(A==B, k=k+1; if(k%5==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