A227631 Array t(n,k): row n consists of the positive integers m for which the least splitter of H(m) and H(m+1) is n, where H denotes harmonic number.

1, 3, 2, 10, 6, 5, 30, 18, 7, 4, 82, 50, 15, 8, 16, 226, 136, 21, 13, 20, 9, 615, 372, 42, 23, 24, 12, 14, 1673, 1014, 59, 38, 36, 25, 19, 44, 4549, 2758, 115, 64, 45, 35, 22, 56, 17, 12366, 7500, 161, 106, 55, 70, 26, 73, 33, 11, 33616, 20389, 315, 175, 67

Offset: 1

Author

Clark Kimberling, Jul 18 2013

Keywords

• nonn
• tabl
• frac

Comments

Suppose that x < y. The least splitter of x and y is introduced here as the least positive integer d such that x <= c/d < y for some integer c; the number c/d is called the least splitting rational of x and y. Conjecture: every row of the array in A227631 is infinite, and every positive integer occurs exactly once. Let r be the limiting ratio of consecutive terms of row 1; is r = e?

For any reduced fraction c/d with c sufficiently large, there exists an (H(m),H(m+1)) pair bracketing it to within 1/LCM(1,2,3,...,d), so d is the least splitter for all such pairs, and every row is infinite. Since each (H(m),H(m+1)) pair is assigned to a single row, each positive integer m occurs exactly once by construction. Since t(1,k) = A002387(k) - 1 for all k >= 1, r = e is indeed the limiting ratio for row 1. - Matthew House, Aug 14 2024

Examples

			Northwest corner of the array:
  1 ... 3 ... 10 ... 30 ... 82 ... 226
  2 ... 6 ... 18 ... 50 ... 136 .. 372
  5 ... 7 ... 15 ... 21 ... 42 ... 59
  4 ... 8 ... 13 ... 23 ... 38 ... 64
  16 .. 20 .. 24 ... 36 ... 45 ... 55
  9 ... 12 .. 25 ... 35 ... 70 ... 97
  14 .. 19 .. 22 ... 26 ... 34 ... 40
t(2,1) = 2 matches 1 + 1/2 <= 3/2 < 1 + 1/2 + 1/3;
similarly, t(2,2) = 6 matches H(6) <= 5/2 < H(7) and t(2,3) = 18 matches H(18) <= 7/2 < H(19).
		

Links

• Matthew House, Table of n, a(n) for n = 1..10011 (rows 1..141).

Crossrefs

Cf. A227629, A227630.

Programs

Mathematica

h[n_] := h[n] = HarmonicNumber[n]; r[x_, y_] := Module[{c, d}, d = NestWhile[#1 + 1 &, 1, ! (c = Ceiling[#1 x - 1]) < Ceiling[#1 y] - 1 &]; (c + 1)/d]; t = Table[r[h[n], h[n + 1]], {n, 1, 40000}];
d = Denominator[t]; u[n_] := Flatten[Position[d, n]]; TableForm[Table[u[n], {n, 1, 50}]]  (* A227631  *)
r1[n_, k_] := u[n][[k]]; z = 11;  v = Flatten[Table[r1[n - k + 1, k], {n, z}, {k, n, 1, -1}]]  (*  A227631 sequence *)  (* Peter J. C. Moses, Jul 15 2013 *)