A070000 Let B(n) denote the number of distinct norms <=n in the square lattice and let B_3(n) denote the number of distinct norms <=n in the hexagonal lattice; sequence gives B(n) - B3(n).

0, 1, 0, 0, 1, 1, 0, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 1, 2, 2, 1, 2, 2, 3, 3, 3, 3, 3, 2, 3, 4, 4, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 5, 5, 5, 5, 4, 5, 5, 5, 5, 5, 4, 4, 5, 5, 4, 5, 5, 5, 5, 6, 6, 7, 6, 5, 5, 5, 4, 5, 5, 6, 6, 5, 6, 6, 6, 6, 7, 8, 7, 7, 6, 6, 6, 6, 6, 7, 7

Offset

0,10

Comments

It was conjectured by Schmutz Schaller that B(x)>=B_3(x) for every x; this was proved in the reference given.

Links

Table of n, a(n) for n=0..98.

P. Moree and H. J. J. te Riele, The hexagonal versus the square lattice, arXiv:math/0204332 [math.NT], 2002.

P. Moree and H. J. J. te Riele, The hexagonal versus the square lattice, Math. Comp. 73 (2004), no. 245, 451-473.

Maple

for j from 0 to 200 do; a[j] := 0; b[j] := 0; end do: for i from 0 to 15 do; for j from 0 to 15 do; u := i*i+j*j; v := i*i+3*j*j; if u<201 then a[u] := 1; end if; if v<201 then b[v] := 1; end if; end do; end do: u := 0: for j from 0 to 200 do; u := u+a[j]-b[j]; print(j, u); end do:

Mathematica

max = 99; rmax = Floor[Sqrt[max]]; ClearAll[a, b]; a[_] = 0; b[_] = 0; Do[u = i*i + j*j; v = i*i + 3*j*j; If[u < max, a[u] = 1]; If[v < max, b[v] = 1], {i, 0, rmax}, {j, 0, rmax}]; Accumulate[ Array[a, max] - Array[b, max]] (* Jean-François Alcover, Oct 18 2011, after Maple *)

Crossrefs

Sequence in context: A335230 A300752 A300751 * A227736 A228528 A366040

Adjacent sequences: A069997 A069998 A069999 * A070001 A070002 A070003

Keyword

nonn,nice,easy

Author

P. Moree (moree(AT)mpim-bonn.mpg.de), May 03 2002

Status

approved