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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.
4
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).
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}
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 14 2005
STATUS
approved