cp's OEIS Frontend

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

%I A364609 #19 Sep 08 2023 07:09:22
%S A364609 1,5,9,13,18,22,26,30,34,38,42,46,50,55,59,63,67,71,75,79,83,87,92,96,
%T A364609 100,104,108,112,116,120,124,129,133,137,141,145,149,153,157,161,166,
%U A364609 170,174,178,182,186,190,194,198,203,207,211,215,219,223,227,231
%N A364609 a(n) = greatest integer k such that 1/n + 1/(n + 1) + ... + 1/k < sqrt(2).
%C A364609 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.
%e A364609 a(3) = 9 because 1/3 + 1/4 + ... + 1/9 < sqrt(2) < 1/3 + 1/4 + ... + 1/10.
%t A364609 r = Sqrt[2]; h[n_] := HarmonicNumber[n];
%t A364609 a[n_] : = Select[Range[500], h[#] <= r + h[n - 1] < h[# + 1] & ]
%t A364609 Flatten[Table[a[n], {n, 1, 70}]]
%o A364609 (Python)
%o A364609 from itertools import accumulate, count
%o A364609 from fractions import Fraction
%o A364609 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
%o A364609 (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
%Y A364609 Cf. A001620, A136616, A357923, A363993.
%K A364609 nonn
%O A364609 1,2
%A A364609 _Clark Kimberling_, Sep 06 2023