A112560 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 2 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

1, 4, 13, 28, 61, 88, 133, 208, 313, 364, 541, 724, 853, 1048, 1261, 1564, 1993, 2104, 2581, 3028, 3553, 3904, 4621, 5368, 5893, 6544, 7141, 8104, 9373, 9904, 11113, 12088, 13333, 14428, 15433, 17368, 19021, 20188, 21733, 23944, 25261, 27304

Offset: 0

Author

Paul D. Hanna, Oct 14 2005

Keywords

• nonn

Comments

Formula: a(n) = 1 + [...[[[[n*2/1]3/2]5/4]6/5]...(k+1)/k]... where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 3); thus the product will eventually reach a maximum value of a(n).

Examples

			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 2, repeat; giving
1,4,7,10,13,16,19,22,25,28,31,34,37,40,...
Step 2: keep 2 terms, remove the next 2, repeat; giving
1,4,13,16,25,28,37,40,49,52,61,64,73,76,...
Step 3: keep 3 terms, remove the next 2, repeat; giving
1,4,13,28,37,40,61,64,73,88,97,100,121,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2)=1+[[[[[[[(2)*2/1]3/2]5/4]6/5]8/7]9/8]11/10]12/11]=13.
		

Crossrefs

Cf. A073360, A112561, A112562, A112563, A112564, A112565, A112566, A112567, A112568, A112569.

Programs

Mathematica

Table[1 + First@FixedPoint[{Floor[#[[1]]*(#[[2]] + 1)/#[[2]]],
      If[Mod[#[[2]] + 1, 3] == 0, #[[2]] + 2, #[[2]] + 1]} &, {n, 1},
SameTest -> (#1[[1]] == #2[[1]] &)], {n, 0, 30}] (* Birkas Gyorgy, Mar 07 2011 *)

PARI

{a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%3==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}

Formula

a(n) = 1 + 3*A073360(n).