cp's OEIS Frontend

A152009 (L)-sieve transform of {1,4,7,10,...,3n-2,...} (A016777).

%I A152009 #18 Jun 02 2025 01:12:47
%S A152009 1,3,6,10,16,25,39,60,91,138,208,313,471,708,1063,1596,2395,3594,5392,
%T A152009 8089,12135,18204,27307,40962,61444,92167
%N A152009 (L)-sieve transform of {1,4,7,10,...,3n-2,...} (A016777).
%C A152009 The (L)-sieve transform of the sequence {a(n)} of positive integers is defined as follows: Denote the sequence of natural numbers by N. Remove the first term of N, which we denote by s(1) and then from the resulting sequence delete all terms whose index is a term of {a(n)}, to obtain the sequence N'.
%C A152009 Then remove the first term of N', denoted by s(2) and then from the resulting sequence delete all terms whose index is a term of {a(n)}, to obtain N''. Repeat this process indefinitely to obtain the transform LST({a(n)}) = {s(1), s(2),...}, the sequence of initial terms removed at each stage.
%C A152009 The (L)-sieve transform is quite different from the transform introduced by _N. J. A. Sloane_ in A099361 and used by T. D. Noe in A100424 - A100426 and seems to lead to more interesting results and relationships among sequences. An interesting property of the (L)-sieve transform is that the (L)-sieve transform of the sequence {1,3,6,10,...,n(n+1)/2,...} of triangular numbers is again the triangular numbers. Another (conjectured) connection with the triangular numbers is given in the following.
%C A152009 Conjecture. Let x(0) be a random sequence of positive integers and, for n>0, let x(n)=S[x(n-1)], where S is the (L)-sieve transform. Then the limit of {x(n)} as n goes to infinity is the sequence of triangular numbers {1,3,6,10,...,n(n+1)/2,...}.
%C A152009 Illustration of the conjecture:
%C A152009 x(0)={3,8,12,14,18,22,25,31,34,39,42,45,...} (A random initial sequence.)
%C A152009 x(1)={1,2,3,5,7,10,14,20,28,38,51,69,...}
%C A152009 x(2)={1,5,12,20,30,41,53,65,78,91,105,119,...}
%C A152009 x(3)={1,3,5,8,11,15,19,24,29,35,41,48,...}
%C A152009 x(4)={1,3,7,13,21,31,43,56,71,88,107,127,...}
%C A152009 x(5)={1,3,6,10,15,20,26,33,40,48,56,65,...}
%C A152009 x(6)={1,3,6,10,15,22,30,39,50,62,75,90,...}
%C A152009 x(7)={1,3,6,10,15,21,28,36,45,55,66,78,...} ...
%C A152009 t={1,3,6,10,15,21,28,36,45,55,66,78,...} (Triangular numbers)
%F A152009 It appears that {a(n)} is given by a(n)=floor[(3*a(n-1)+3)/2], with a(1)=1.
%o A152009 (Maxima)
%o A152009 a[1]:1$
%o A152009 a[n]:=floor((3*a[n-1]+3)/2)$
%o A152009 A152009(n):=a[n];
%o A152009 makelist(A152009(n),n,1,30); /* _Martin Ettl_, Oct 31 2012 */
%K A152009 nonn
%O A152009 1,2
%A A152009 _John W. Layman_, Nov 19 2008