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Let h(2*n, 1) = 2*n and h(2*n, m) = h(2*n, m-1) + 2 * d(2*n, m) for m > 1, where d(2*n, m) = (least multiple of m not less than h(2*n, m-1)) - h(2*n, m-1). Then d(2*n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2*n, 2*m+1) for large m.
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%I #15 May 02 2023 03:02:38

%S 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,

%T 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,

%U 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

%N Let h(2*n, 1) = 2*n and h(2*n, m) = h(2*n, m-1) + 2 * d(2*n, m) for m > 1, where d(2*n, m) = (least multiple of m not less than h(2*n, m-1)) - h(2*n, m-1). Then d(2*n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2*n, 2*m+1) for large m.

%C Sequence yielding an ordering of N*N derived from a family of recurrences.

%C For any integer k define h(k,1)=k and for m>1 define h(k,m)=h(k,m-1)+2*((-h(k,m-1)) mod m) where "r mod s" denotes least nonnegative residue of r modulo s [informally, h(k,m) is got by "reflecting" h(k,m-1) in the least multiple of m that is >=h(k,m-1)]. Then for fixed k>=0 there are integers c(k), b(k), m(k) such that for all m>m(k) we have h(k,2*m+1)-h(k,2*m)=2*c(k) and h(k,2*m+2)-h(k,2*m+1)=2*b(k). [In the terms of the function d defined in the Name, c(k) = d(k, 2*m+1) and b(k) = d(k, 2*m+2) for all m>m(k).]

%C For all n we have c(2*n+1)=c(2*n) and b(2*n+1)=1+b(2*n). Moreover b(2*n) is even for all n. The function n->(c(2*n),b(2*n)/2) is a bijection from the nonnegative integers N to N*N. It is "monotone" in the sense that n<=n' whenever c(2*n)<=c(2*n') and b(2*n)<=b(2*n').

%C This sequence is a(n) = c(2*n).

%C The results on which the definition is based are not yet proved, but they are plausible and overwhelmingly supported by numerical evidence.

%C For each fixed m, k->h(k,m) is a bijection Z->Z (this is easy!). However for k<0 the sequence h(k,m) does not have the pseudo-periodic property we have used in defining c(k) and b(k).

%C m(k) appears to be O(sqrt k).

%F It appears that a(n) = A319573(A252448(2*n+1)). [discovered by Sequence Machine] Also, it appears that b(2*n) = A319572(A252448(2*n)), where b is described above. - _Andrey Zabolotskiy_, Apr 28 2023

%e h(18,m) for m>=1 goes 18,18,18,22,28,32,38,42,48...so we can take m(18) = 2, then 2*c(18)=28-22=38-32=48-42=...=6, so a(9) = c(18) = 3 and similarly b(18) = 2.

%t d[h_, m_] := (Ceiling[#] - # &[h/m]) m;

%t a[n_] := Module[{h = 2 n, b, c, m = 1},

%t While[m <= Max[5, n], (*condition is conjectural*)

%t m++; b = d[h, m]; h += 2 b;

%t m++; c = d[h, m]; h += 2 c];

%t c];

%t Table[a[n], {n, 0, 93}]; (* _Andrey Zabolotskiy_, Apr 29 2023 *)

%Y Cf. A319571, A319572, A319573, A252448.

%K nonn

%O 0,3

%A _Alex Abercrombie_, Feb 15 2008

%E Edited by _Andrey Zabolotskiy_, Apr 29 2023