A364609 a(n) = greatest integer k such that 1/n + 1/(n + 1) + ... + 1/k < sqrt(2).

1, 5, 9, 13, 18, 22, 26, 30, 34, 38, 42, 46, 50, 55, 59, 63, 67, 71, 75, 79, 83, 87, 92, 96, 100, 104, 108, 112, 116, 120, 124, 129, 133, 137, 141, 145, 149, 153, 157, 161, 166, 170, 174, 178, 182, 186, 190, 194, 198, 203, 207, 211, 215, 219, 223, 227, 231

Offset

1,2

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 = sqrt(2), the two numbers are 4 and 5.

Links

Table of n, a(n) for n=1..57.

Example

a(3) = 9 because 1/3 + 1/4 + ... + 1/9 < sqrt(2) < 1/3 + 1/4 + ... + 1/10.

Mathematica

r = Sqrt[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 A364609(n): return next(x[0]+n-1 for x in enumerate(accumulate(Fraction(1, k) for k in count(n))) if x[1]**2 >= 2) # Chai Wah Wu, Sep 07 2023
(PARI) a(n) = my(k=0); while (sum(i=n, n+k, 1/i)^2 < 2, k++); n+k-1; \\ Michel Marcus, Sep 08 2023

Crossrefs

Cf. A001620, A136616, A357923, A363993.

Sequence in context: A314703 A314704 A314705 * A314706 A081835 A314707

Adjacent sequences: A364606 A364607 A364608 * A364610 A364611 A364612

Keyword

nonn

Author

Clark Kimberling, Sep 06 2023

Status

approved