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

0,2

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).

Links

Table of n, a(n) for n=0..41.

Formula

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

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 *)

Prog

(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}

Crossrefs

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

Sequence in context: A054968 A087035 A212578 * A009561 A155266 A155413

Adjacent sequences: A112557 A112558 A112559 * A112561 A112562 A112563

Keyword

nonn

Author

Paul D. Hanna, Oct 14 2005

Status

approved