Integer Sequences Partition Positive Integers

[Leroy Quet]

I just posted the message below to sci.math.  I am crossposting to 
seq.fan, however, because it directly involves sequences of positive 
integers.
(It is along the same lines as a message I posted here a couple weeks 
ago.)

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Here is a result which is pretty much the same result mentioned in

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=ISO-8859-1&safe=off&t
hreadm=b4be2fdf.0404121104.251cda6d%40posting.google.com&rnum=18&prev=

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=ISO-8859-1&safe=off&t
hreadm=b4be2fdf.0404121120.28bba1c2%40posting.google.com&rnum=17&prev=

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=ISO-8859-1&safe=off&t
hreadm=b4be2fdf.0404151306.22187e2d%40posting.google.com&rnum=16&prev=

but involves integer sequences, not real functions.


Let {a(k)} be any sequence of positive integers, where
a(k+1) >= a(k) for each k.
(a(1) is the sequence's first term.)


For m = a positive integer,
let b(k) be the number of elements
among the first m terms of {a(k)} which are <= k.


So, every positive integer from 1 to (a(m)+m-1) occurs exactly once
either in the sequence

{a(k)+k-1}, for 1<=k<=m,

or in

{b(k)+k}, for 1<=k<= a(m)-1,

but not in both sequences.
(And, for the range of k's, only these integers occur.)


For example:
Let a(k) = p(k), the k_th prime.

Let b(k) = pi(k), the number of primes <= k.

So, if m is infinity, we have the sequences:

{p(k)+k-1}: 2, 4, 7, 10, 15, 18, 23,...
{pi(k)+k}:  1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16,...

thanks,
Leroy Quet