login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A227965
a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.
5
11, 53, 249, 1164, 5435, 25371, 118428, 552798, 2580343, 12044484, 56221045, 262427666, 1224955522, 5717827134, 26689578960, 124581175389, 581517950673, 2714399875409, 12670230858892, 59141894115145, 276061555506087, 1288595564424512, 6014885070144844
OFFSET
1,1
COMMENTS
Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?
For A227965, (x,y) = (1,2); H(a(n)) - H(a(n-1)) approaches a limit 1.540684... given by A227966, and a(n)/a(n-1) approaches a limit 4.6677834... given by A227967. It is unknown whether the sequence a(n) is linearly recurrent.
LINKS
EXAMPLE
The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...
MATHEMATICA
z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227965 *)
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]] (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]] (* A227967 *)
(* A227965, Peter J. C. Moses, Jul 12 2013*)
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Clark Kimberling, Aug 01 2013
STATUS
approved