A389596 Harmonic-mean index of primes: a(1) = 0 and thereafter a(n) = floor( n / (2*H(n)) ), where H(n) = Sum_{k=1..n} 1/prime(k).

0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18

Offset

1,6

Comments

For n >= 2, f(n) = n/(2*H(n)) is strictly increasing and never an integer, hence a(n) is nondecreasing with unit jumps. Indeed, f(n+1) - f(n) = (p*H(n) - n) / (2*H(n)*(p*H(n) + 1)), where p = prime(n+1), so 0 < f(n+1) - f(n) < 1/(2*H(n)) < 1 (since H(2) = 5/6 > 1/2 and H is increasing).

Let P_n = A002110(n) (primorial) and S_n = Sum_{k=1..n} P_n/p_k = A024451(n). Then gcd(S_n, P_n) = 1 and S_n is odd. Consequently f(n) = n*P_n/(2*S_n) can be an integer only at n = 1; therefore the only tie case is n = 1, giving a(1) = 0, and for n >= 2 we may simply use floor.

Define t(m) = min{ n : a(n) = m } and L(m) = t(m+1) - t(m). Since f is strictly increasing and non-integral (n >= 2), t(m) = min{ n : f(n) >= m }. Asymptotically L(m) = 2*log log m + O(1) = Theta(log log m). From Rosser-Schoenfeld bounds one obtains effective constants c1, c2 and m0 with c1 + 2*log log m <= L(m) <= c2 + 2*log log m for all m >= m0 (e.g., a conservative choice c1 = -10, c2 = +10, m0 = 16).

a(n) = n / (2*(log log n + B1)) + O( n/(log n*log log n) ), using prime(n) ~ n*(log n + log log n - 1 + ...) and H(n) = log log n + B1 + (log log n - 1)/log n + O( (log log n)^2/(log n)^2 ), where B1 = A077761.

t(m) ~ 2*m*(log log m + B1), and L(m) ~ 2*log log m + O(1).

References

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford Univ. Press, 2008, Chap. XXII; see Sec. 22.18 "Products of k prime factors", pp. 383-386 (6th ed. pagination).

Links

Pablo Cadena-Urzúa, Table of n, a(n) for n = 1..10000

J. Barkley Rosser and Lowell Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 (1962), 64-94.

Formula

a(n) = floor( n*A002110(n) / (2*A024451(n)) ), for n >= 2.

Example

n=1: primes {2}, H(1) = 1/2, f(1) = 1, a(1) = 0.
n=5: primes {2,3,5,7,11}, H(5) = 2927/2310, f(5) = 5775/2927 = 1.973..., a(5) = 1.

Mathematica

primorial[n_] := Product[Prime[i], {i, n}];
a024451[n_] := Numerator @ Sum[1/Prime[i], {i, n}];
a[n_]:=Floor[n*primorial[n]/(2*a024451[n])]; a[1]=0; Array[a, 75] (* James C. McMahon, Nov 13 2025 *)

Prog

(Python)
from sympy import prime
def a_values(N: int):
    if N < 1:
        return []
    P = 2
    S = 1
    out = [0]
    if N == 1:
        return out
    for n in range(2, N + 1):
        p = prime(n)
        P_prev = P
        P *= p
        S = p*S + P_prev
        out.append((n*P) // (2*S))
    return out
(Python)
from fractions import Fraction
from sympy import prime
def A389596(n):
    if n == 1: return 0
    return n*(h:=sum(Fraction(1, prime(k)) for k in range(1, n+1))).denominator//(h.numerator<<1) # Chai Wah Wu, Nov 13 2025

Crossrefs

Cf. A000040, A002110, A024451, A034386, A077761, A106830.

Sequence in context: A337635 A140437 A226763 * A050500 A076885 A274024

Adjacent sequences: A389593 A389594 A389595 * A389597 A389598 A389599

Keyword

nonn

Author

Pablo Cadena-Urzúa, Oct 10 2025

Status

approved