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A112561 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 3 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

%I A112561 #4 Mar 30 2012 18:36:51
%S A112561 1,5,21,61,125,261,421,605,1101,1681,2525,2781,4201,5645,6741,9541,
%T A112561 11765,13701,17641,21305,27981,29401,37265,43521,51541,59945,65781,
%U A112561 78121,89345,99981,121381,124445,144321,173041,189965,212361,229381
%N A112561 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 3 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.
%C A112561 Formula: a(n) = 1 + [..[[[[n*2/1]3/2]4/3]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 4); thus the product will eventually reach a maximum value of a(n).
%F A112561 a(n) = 1 + 4*A073361(n).
%e A112561 Sieve starts with the natural numbers:
%e A112561 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15...
%e A112561 Step 1: keep 1 term, remove the next 3, repeat; giving
%e A112561 1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,...
%e A112561 Step 2: keep 2 terms, remove the next 3, repeat; giving
%e A112561 1,5,21,25,41,45,61,65,81,85,101,105,121,125,141,...
%e A112561 Step 3: keep 3 terms, remove the next 3, repeat; giving
%e A112561 1,5,21,61,65,81,121,125,141,181,185,201,241,245,261,...
%e A112561 Continuing in this way, we obtain this sequence.
%e A112561 Using the floor function product formula:
%e A112561 a(2)=1+[..[(2)*2/1]*3/2]*4/3]*6/5]*7/6]*8/7]*10/9]*11/10]*
%e A112561 12/11]*14/13]*15/14]*16/15]*18/17]*19/18]*20/19] = 21.
%o A112561 (PARI) {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%4==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}
%Y A112561 Cf. A073361, A112560, A112562, A112563, A112564, A112565, A112566, A112567, A112568, A112569.
%K A112561 nonn
%O A112561 0,2
%A A112561 _Paul D. Hanna_, Oct 14 2005