A224820 Array r(n,m), where r(n,1) = n; r(n,2) = least k such that H(k) - H(n) > 1/n; and for m > 2, r(n,m) = least k such that H(k)-H(r(n,m-1)) > H(r(n,m-1)) - H(r(n,m-2)), where H = harmonic number.

1, 4, 2, 13, 4, 3, 40, 8, 5, 4, 121, 16, 9, 6, 5, 364, 32, 16, 9, 7, 6, 1093, 64, 29, 14, 10, 8, 7, 3280, 128, 53, 22, 15, 11, 9, 8, 9841, 256, 97, 35, 23, 16, 12, 10, 9, 29524, 512, 178, 56, 36, 24, 16, 13, 11, 10, 88573, 1024, 327, 90, 57, 36, 22, 17, 14, 12, 11

Offset: 1

Author

Clark Kimberling, Jul 21 2013

Keywords

• nonn
• tabl

Comments

For every n, the sequence H(r(n,m)) - H(r(n,m-1)) converges as m -> oo. Which row-sequences are linearly recurrent? Is r(4,m) = 1 + F(m+3), where F = A000045 (Fibonacci numbers)?

More generally, suppose that x and y are positive integers and that x <= y. Let c(1) = y and c(2) = least k such that H(k) - H(y) > H(y) - H(x); for n > 2, let c(n) = least k such that H(k) - H(c(n-1)) > H(c(n-1)) - H(c(n-2)). Thus the Egyptian fractions for m >= x are partitioned, and 1/x + ... + 1/c(1) < 1/(c(1)+1) + ... + 1/(c(2)) < 1/(c(2)+1) + ... + 1/(c(3)) < ... The sequences H(c(n))-H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

Examples

			Northwest corner:
       m=1  m=2  m=3  m=4  m=5  m=6   m=7   m=8
  n=1:   1,   4,  13,  40, 121, 364, 1093, 3280
  n=2:   2,   4,   8,  16,  32,  64,  128,  256
  n=3:   3,   5,   9,  16,  29,  53,   97,  178
  n=4:   4,   6,   9,  14,  22,  35,   56,   90
  n=5:   5,   7,  10,  15,  23,  36,   57,   91
  n=6:   6,   8,  11,  16,  24,  36,   54,   81
  n=7:   7,   9,  12,  16,  22,  31,   44,   63
  n=8:   8,  10,  13,  17,  23,  32,   45,   64
The chain indicated by row n=4 is
1/4 < 1/5 + 1/6 < 1/7 + 1/8 + 1/9 < 1/10 + ... + 1/14 < ...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1830

Crossrefs

Cf. A225918.

Programs

Mathematica

h[n_] := h[n] = HarmonicNumber[N[n, 300]]; z = 12; Table[s = 0; a[1] = NestWhile[# + 1 &, x + 1, ! (s += 1/#) >= h[x] - h[x - 1] &];   s = 0; a[2] = NestWhile[# + 1 &, a[1] + 1, ! (s += 1/#) >= h[a[1]] - h[x] &]; Do[test = h[a[t - 1]] - h[a[t - 2]] + h[a[t - 1]]; s = 0; a[t] = Floor[x /. FindRoot[h[x] == test, {x, a[t - 1]}, WorkingPrecision -> 100]] + 1, {t, 3, z}]; Flatten[{x, Map[a, Range[z]]}], {x, 1, z}] // TableForm (* A224820 array *)
t = Flatten[Table[%[[n - k + 1]][[k]], {n, z}, {k, n, 1, -1}]]; (* A224820 sequence *) (* Peter J. C. Moses, Jul 20 2013 *)