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A348626 Greedy Egyptian fraction representation of 1 with square denominators.

%I A348626 #46 Apr 21 2025 16:54:59
%S A348626 2,2,2,3,3,7,12,49,340,6153,362275,234314697,4303312007019,
%T A348626 8064823505928103487,21034270897045389505182033301,
%U A348626 13184627067084215135862894820778146400791573,36011454158212923548860166370685543204871921069986403871775848271,6820216143160044256325325882329406136711110111012515344838677137010956148075846307036940303634819
%N A348626 Greedy Egyptian fraction representation of 1 with square denominators.
%C A348626 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]
%C A348626 Comment from _David desJardins_, Apr 20 2025 (Start)
%C A348626 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
%C A348626    0 < g(n) < 1/(a(n)-1)^2 - 1/a(n)^2 \approx 2/a(n)^3.
%C A348626 So a(n)^3/a(n+1)^2 should be roughly uniform on [0,2].
%C A348626 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.
%C A348626 So L(n+1) \approx (3/2)*L(n) + (1/2)*(e(n+1)-ln(2)).
%C A348626 This gives us the conjecture that L(n) = C * (3/2)^n * (1+o(1)), as n -> oo.
%C A348626 The plot of L(n)*(2/3)^n (see link) shows that the conjecture is plausible, with C \approx 0.15113.
%C A348626 (End)
%H A348626 David desJardins, <a href="/A348626/b348626.txt">Table of n, a(n) for n = 1..24</a>
%H A348626 David desJardins, <a href="/A348626/a348626_1.pdf">Plot of log(a(n))*(2/3)^n vs n</a>
%H A348626 David desJardins, <a href="/A348626/a348626.txt">Comma-separated list of first 40 terms</a>
%o A348626 (PARI) s=1; for(n=1,20, t=sqrtint(floor(1/s))+1; s-=1/t^2; print1(t,", "));
%o A348626 (Python)
%o A348626 from math import isqrt
%o A348626 from fractions import Fraction
%o A348626 def A348626List():
%o A348626     s = Fraction(1, 1)
%o A348626     while True:
%o A348626         t = isqrt(1 // s) + 1
%o A348626         yield t
%o A348626         s -= Fraction(1, t * t)
%o A348626 a = A348626List()
%o A348626 print([next(a) for _ in range(18)])  # _Peter Luschny_, Oct 26 2021
%Y A348626 Cf. A000058, A348625.
%K A348626 nonn
%O A348626 1,1
%A A348626 _Max Alekseyev_, Oct 25 2021