A259578 Reciprocity array of 2; rectangular, read by antidiagonals.

2, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 4, 3, 4, 2, 2, 4, 5, 5, 4, 2, 2, 5, 6, 6, 6, 5, 2, 2, 5, 6, 8, 8, 6, 5, 2, 2, 6, 8, 10, 10, 10, 8, 6, 2, 2, 6, 9, 11, 12, 12, 11, 9, 6, 2, 2, 7, 9, 12, 14, 15, 14, 12, 9, 7, 2, 2, 7, 11, 14, 16, 17, 17, 16, 14, 11, 7, 2, 2, 8

Offset: 1

Author

Clark Kimberling, Jul 17 2015

Keywords

• nonn
• easy
• tabl

Comments

The "reciprocity law" that Sum_{k=0..m} [(n*k+x)/m] = Sum_{k=0..n} [(m*k+x)/n] where x is a real number and m and n are positive integers, is proved in Section 3.5 of Concrete Mathematics (see References). See A259572 for a guide to related sequences.

Examples

			Northwest corner:
  2   2   2   2   2   2   2   2   2   2
  2   3   3   4   4   5   5   6   6   7
  2   3   3   5   6   6   8   9   9   11
  2   4   5   6   8   10  11  12  14  16
  2   4   6   8   10  12  14  16  18  20
  2   5   6   10  12  15  17  20  21  25
		

References

• R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989, pages 90-94.

Links

• Clark Kimberling, Antidiagonals n=1..60, flattened

Crossrefs

Cf. A259572, A259579, A259580.

Programs

Mathematica

x = 2; s[m_, n_] := Sum[Floor[(n*k + x)/m], {k, 0, m - 1}];
TableForm[ Table[s[m, n], {m, 1, 15}, {n, 1, 15}]]
u = Table[s[n - k + 1, k], {n, 15}, {k, n, 1, -1}] // Flatten

Formula

T(m,n) = Sum_{k=0..m-1} [(n*k+x)/m] = Sum_{k=0..n-1} [(m*k+x)/n], where x = 2 and [ ] = floor. Note that if [x] = [y], then [(n*k+x)/m] = [(n*k+y)/m], so that the reciprocity arrays for x and y are identical.