0,2

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).

a(n) = 1 + 6*A073363(n).

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.

(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}

Cf. A073363, A112560, A112561, A112562, A112564, A112565, A112566, A112567, A112568, A112569.

Adjacent sequences: A112560 A112561 A112562 this_sequence A112564 A112565 A112566

Sequence in context: A031914 A134154 A114352 this_sequence A143569 A027176 A079925

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Oct 14 2005