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A112562 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 4 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

%I A112562 #4 Mar 30 2012 18:36:51
%S A112562 1,6,31,96,241,546,1171,1776,2761,5046,7591,11856,19021,20706,31711,
%T A112562 44016,60481,71946,92191,120216,138601,181986,226831,302496,379381,
%U A112562 400686,487831,574656,704461,831606,1029631,1092936,1333321,1462146
%N A112562 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 4 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.
%C A112562 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 5); thus the product will eventually reach a maximum value of a(n).
%F A112562 a(n) = 1 + 5*A073362(n).
%e A112562 Sieve starts with the natural numbers:
%e A112562 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,...
%e A112562 Step 1: keep 1 term, remove the next 4, repeat; giving
%e A112562 1,6,11,16,21,26,31,36,41,46,51,56,61,66,...
%e A112562 Step 2: keep 2 terms, remove the next 4, repeat; giving
%e A112562 1,6,31,36,61,66,91,96,121,126,151,156,...
%e A112562 Step 3: keep 3 terms, remove the next 4, repeat; giving
%e A112562 1,6,31,96,121,126,211,216,241,306,331,...
%e A112562 Continuing in this way, we obtain this sequence.
%e A112562 Using the floor function product formula:
%e A112562 a(2) = 1 + [..[(2)*2/1]*3/2]*4/3]*6/5]*7/6]*8/7]*10/9]*
%e A112562 11/10]*12/11]*14/13]*15/14]*16/15]*18/17]*19/18]*20/19] = 21.
%o A112562 (PARI) {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%5==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}
%Y A112562 Cf. A073362, A112560, A112561, A112563, A112564, A112565, A112566, A112567, A112568, A112569.
%K A112562 nonn
%O A112562 0,2
%A A112562 _Paul D. Hanna_, Oct 14 2005