OFFSET
1,2
COMMENTS
It appears that every positive integer k occurs exactly once or exactly twice as a solution of H(n) - H(n+k) < g-1 < H(n) - H(n+k-1) as n runs through the positive integers, where H denotes harmonic number, and g denotes the Euler-Mascheroni constant. See A227581.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..400
EXAMPLE
That 5 occurs just once, and 4 and 6 each occur twice, corresponds to these inequalities:
H(6) - H(6 + 4) < g-1 < H(6) - H(6 + 3),
H(7) - H(7 + 4) < g-1 < H(7) - H(7 + 3),
H(8) - H(8 + 5) < g-1 < H(8) - H(8 + 4),
H(9) - H(9 + 6) < g-1 < H(9) - H(9 + 5),
H(10) - H(10 + 6) < g-1 < H(10) - H(10 + 5).
MATHEMATICA
z = 1500; r[n_] := r[n] = Module[{Nn = N[n, 50]}, NestWhile[# + 1 &, Floor[(1 + n)/2], ! HarmonicNumber[1] + HarmonicNumber[Nn] - HarmonicNumber[Nn + #] < EulerGamma &]]; u[k_] := Length[Split[Table[r[n], {n, z}]][[k]]]; t = Table[u[k], {k, 1, z/2}]; Flatten[Position[t, 1]] (* A227586 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 17 2013
STATUS
approved