A363993 a(n) = greatest integer k such that 1/n + 1/(n + 1) +...+ 1/k < 2.

3, 10, 18, 25, 32, 40, 47, 54, 62, 69, 77, 84, 91, 99, 106, 114, 121, 128, 136, 143, 150, 158, 165, 173, 180, 187, 195, 202, 210, 217, 224, 232, 239, 247, 254, 261, 269, 276, 283, 291, 298, 306, 313, 320, 328, 335, 343, 350, 357, 365, 372, 380, 387, 394, 402

Offset

1,1

Comments

In general, if r > 0 and n > 1, then a(n) is the number k such that h(k) <= r + h(n-1) < h(k+1), where h(m) = m-th harmonic number. Since h(n) is approximately g + log(n+1/2), where g = Euler-Mascheroni constant (A001620), it is easy to prove that a(n) or a(n)-1 is the number floor[n*e^r-(1+e^r)/2]. Thus, the difference sequence of (a(n)) has at most two distinct numbers. For r = 2, the two numbers are 7 and 8.

Links

Table of n, a(n) for n=1..55.

Example

a(2) = 10 because 1/2 + 1/3 +...+ 1/10 < 2 < 1/2 + 1/3 +...+ 1/11.

Mathematica

r = 2; h[n_] := HarmonicNumber[n];
a[n_] : = Select[Range[500], h[#] <= r + h[n - 1] < h[# + 1] & ]
Flatten[Table[a[n], {n, 1, 70}]]

Prog

(Python)
from itertools import accumulate, count
from fractions import Fraction
def A363993(n): return next(x[0]+n-1 for x in enumerate(accumulate(Fraction(1, k) for k in count(n))) if x[1] >= 2) # Chai Wah Wu, Sep 07 2023
(PARI) a(n) = my(k=0); while (sum(i=n, n+k, 1/i) < 2, k++); n+k-1; \\ Michel Marcus, Sep 08 2023

Crossrefs

Cf. A001620, A136616, A364609.

Sequence in context: A035488 A063211 A063111 * A031063 A051938 A171834

Adjacent sequences: A363990 A363991 A363992 * A363994 A363995 A363996

Keyword

nonn

Author

Clark Kimberling, Sep 06 2023

Status

approved