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%I #20 Sep 10 2023 09:09:25
%S 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,
%T 8,7,3280,128,53,22,15,11,9,8,9841,256,97,35,23,16,12,10,9,29524,512,
%U 178,56,36,24,16,13,11,10,88573,1024,327,90,57,36,22,17,14,12,11
%N 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.
%C 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)?
%C 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?
%H Clark Kimberling, <a href="/A224820/b224820.txt">Table of n, a(n) for n = 1..1830</a>
%e Northwest corner:
%e m=1 m=2 m=3 m=4 m=5 m=6 m=7 m=8
%e n=1: 1, 4, 13, 40, 121, 364, 1093, 3280
%e n=2: 2, 4, 8, 16, 32, 64, 128, 256
%e n=3: 3, 5, 9, 16, 29, 53, 97, 178
%e n=4: 4, 6, 9, 14, 22, 35, 56, 90
%e n=5: 5, 7, 10, 15, 23, 36, 57, 91
%e n=6: 6, 8, 11, 16, 24, 36, 54, 81
%e n=7: 7, 9, 12, 16, 22, 31, 44, 63
%e n=8: 8, 10, 13, 17, 23, 32, 45, 64
%e The chain indicated by row n=4 is
%e 1/4 < 1/5 + 1/6 < 1/7 + 1/8 + 1/9 < 1/10 + ... + 1/14 < ...
%t 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 t = Flatten[Table[%[[n - k + 1]][[k]], {n, z}, {k, n, 1, -1}]]; (* A224820 sequence *) (* _Peter J. C. Moses_, Jul 20 2013 *)
%Y Cf. A225918.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Jul 21 2013
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