A348626 Greedy Egyptian fraction representation of 1 with square denominators.

2, 2, 2, 3, 3, 7, 12, 49, 340, 6153, 362275, 234314697, 4303312007019, 8064823505928103487, 21034270897045389505182033301, 13184627067084215135862894820778146400791573, 36011454158212923548860166370685543204871921069986403871775848271, 6820216143160044256325325882329406136711110111012515344838677137010956148075846307036940303634819

Offset: 1

Author

Max Alekseyev, Oct 25 2021

Keywords

• nonn

Comments

Greedy representation 1 = 1/a(1)^2 + 1/a(2)^2 + ... constructed in a similar way to Sylvester's sequence (A000058). Let s(n) = Sum_{i=1..n} 1/a(i)^2, and let g(n) = 1 - s(n). Each a(n) is taken be the smallest positive integer satisfying s(n) < 1. [Revised by N. J. A. Sloane, Apr 20 2025]

Comment from David desJardins, Apr 20 2025 (Start)

We know that s(n) < 1 and s(n)-1/a(n)^2+1/(a(n)-1)^2 > 1. Then, arguing heuristically, the gap g(n) = 1-s(n) \approx 1/a(n+1)^2. This implies

0 < g(n) < 1/(a(n)-1)^2 - 1/a(n)^2 \approx 2/a(n)^3.

So a(n)^3/a(n+1)^2 should be roughly uniform on [0,2].

Let L(n) = ln(a(n)). Then 3*L(n) - 2*L(n+1) \approx ln(2) - e(n+1), where e(i) has an exponential distribution.

So L(n+1) \approx (3/2)*L(n) + (1/2)*(e(n+1)-ln(2)).

This gives us the conjecture that L(n) = C * (3/2)^n * (1+o(1)), as n -> oo.

The plot of L(n)*(2/3)^n (see link) shows that the conjecture is plausible, with C \approx 0.15113.

(End)

Links

• David desJardins, Table of n, a(n) for n = 1..24
• David desJardins, Plot of log(a(n))*(2/3)^n vs n
• David desJardins, Comma-separated list of first 40 terms

Crossrefs

Cf. A000058, A348625.

Programs

PARI

s=1; for(n=1,20, t=sqrtint(floor(1/s))+1; s-=1/t^2; print1(t,", "));

Python

from math import isqrt
from fractions import Fraction
def A348626List():
    s = Fraction(1, 1)
    while True:
        t = isqrt(1 // s) + 1
        yield t
        s -= Fraction(1, t * t)
a = A348626List()
print([next(a) for  in range(18)])  # _Peter Luschny, Oct 26 2021