

A135317


Sequence yielding an ordering of N*N derived from a family of recurrences. For any integer k define h(k,1)=1 and for n>1 define h(k,n)=h(k,n1)+2*((h(k,n1)mod n)where "r mod s" denotes least nonnegative residue of r modulo s [informally, h(k,n) is got by "reflecting" h(k,n1)in the least multiple of n that is >=h(k,n1)]. Then for fixed k>=0 there are integers a(k), b(k), n(k) such that for all n>n(k) we have h(k,2*n+1)h(k,2*n)=2*a(k)and h(k,2*n+2)h(k,2*n+1)=2*b(k). For all k we have a(2*k+1)=a(2*k) and b(2*k+1)=1+b(2*k). Moreover b(2*k) is even for all k. The function k>(a(2*k),b(2*k)/2) is a bijection from the nonnegative integers N to N*N. It is "monotone" in the sense that k<=k' whenever a(2*k)<=a(2*k') and b(2*k)<=b(2*k'). The sequence given above is a(2*k).


0



0, 1, 2, 0, 1, 2, 3, 4, 0, 3, 4, 5, 1, 2, 3, 6, 0, 1, 4, 5, 6, 2, 5, 6, 7, 8, 0, 7, 8, 9, 3, 4, 5, 1, 2, 3, 6, 7, 10, 4, 7, 8, 0, 1, 2, 9, 10, 11, 5, 8, 9, 12, 6, 7, 10, 11, 12, 8, 9, 10, 0, 3, 4, 13, 1, 2, 5, 6, 11, 3, 4, 9, 14, 0, 1, 10, 11, 12, 2, 7, 8, 13, 5, 6, 9, 12, 13, 7, 10, 11, 14, 15, 16, 14
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OFFSET

0,3


COMMENTS

The results on which the definition is based are not yet proved, but they are plausible and overwhelmingly supported by numerical evidence. I am working on a proof.
For each fixed n, k>h(k,n) is a bijection Z>Z (this is easy!). However for k<0 the sequence h(k,n) does not have the pseudoperiodic property we have used in defining a(k) and b(k).
n(k) appears to be O(sqrt k).


LINKS

Table of n, a(n) for n=0..93.


EXAMPLE

h(18,n) for n>=1 goes 18,18,18,22,28,32,38,42,48...so we can take n(18)=2, then 2*a(18)=2822=3832=4842=...=6, so a(18)=3 and similarly b(18)=2.


CROSSREFS

Sequence in context: A218601 A262678 A238794 * A227179 A115218 A023858
Adjacent sequences: A135314 A135315 A135316 * A135318 A135319 A135320


KEYWORD

nonn,uned


AUTHOR

Alex Abercrombie, Feb 15 2008


STATUS

approved



