A (New?) Sequence Transform
Leroy Quet
qq-quet at mindspring.com
Sat Dec 20 05:06:19 CET 2003
Let {a(k)} be an infinite sequence of positive integers, and where {a}
contains an infinite numbers of terms equal to 1.
Here is a simple(well...) transform which converts {a} into another
infinite sequence {b(k)} of positive integers,
and where {b} also has an infinite number of 1's.
Let {c(k)} be a permutation of the positive integers, where
c(1) = 1, and
c(m+1) = the a(m)_th yet-unpicked positive integer.
(By "yet-unpicked",
I mean that c(m) is not among c(1),c(2),c(3),...c(m-1).)
Let {d(m)} be the inverse of {c}.
(ie. c(d(m)) = m for all m.)
Apply the reverse of the {a}->{c} step to {d} to get {b}.
In other words,
b(m) = the order among + integers
not in {d(1),d(2),d(3),...d(m)}
of d(m+1).
So, this simpler-than-it-must-appear-to-you transform seems like it must
have some uses, but I do not know.
An example of it being used:
Sequence A001222 of EIS:
(from a(2) on)
a -> 1,1,2,1,2,1,3,2,2,1,3,...
c -> 1,2,3,5,4,7,6,10,9,11,8,14,...
d -> 1,2,3,5,4,7,6,11,9,8,10,...
b -> 1,1,2,1,2,1,4,2,1,1,...
(amazing!...)
Aside from if a -> b implies b -> a, what additionally can be said about
this transform on sequences?
thanks,
Leroy
Quet
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