A377570 a(n) = round((H(n) + e^H(n)*log(H(n)) + sigma(n))/2).

1, 3, 5, 7, 8, 12, 12, 16, 17, 21, 19, 28, 23, 29, 30, 35, 30, 42, 34, 46, 43, 46, 41, 61, 48, 55, 55, 65, 53, 76, 57, 74, 68, 73, 71, 94, 69, 82, 81, 100, 77, 106, 81, 103, 101, 100, 89, 129, 97, 116, 107, 122, 102, 136, 114, 139, 121, 127, 114, 169, 118, 137

Offset: 1

Author

Ahmad J. Masad, Nov 01 2024

Keywords

• nonn

Comments

The idea of this sequence is finding the nearest integer to the arithmetic mean between the two sides of the inequality that is equivalent to the Riemann hypothesis. The inequality is sigma(n) < H(n) + e^H(n)*log(H(n)), where H(n) are the harmonic numbers.

Conjecture: For each positive integer k, there exist at least one positive integer m such that there are exactly k terms in this sequence that are equal to m. In the first 10000 terms, each value occurs at most 4 times. However, as n becomes larger, we can see from the scatterplot that the conjecture might be true. - Ahmad J. Masad, Apr 03 2025

Examples

			For n=6, sigma(6)=12 and H(6)=1/1+1/2+1/3+1/4+1/5+1/6=2.45 and (2.45+e^2.45*log(2.45)+12)/2=12.417... and round(12.417...)=12; hence a(6)=12.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A001008/A002805, A057641.

Programs

Mathematica

f[x_] := x + Exp[x] * Log[x]; a[n_] := Round[(f[HarmonicNumber[n]] + DivisorSigma[1, n])/2]; Array[a, 100] (* Amiram Eldar, Nov 01 2024 *)