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A224820
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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.
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11
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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
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OFFSET
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1,2
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COMMENTS
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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?
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LINKS
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EXAMPLE
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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 < ...
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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