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A295391 a(1) = 2; a(n+1) = 1 + (a(n)^2 - a(n))*(1+1/n) for n >= 1.

%I A295391 #20 Oct 20 2024 23:45:32
%S A295391 2,5,31,1241,1923551,4440055831261,22999778415495431257188671,
%T A295391 604559779613588034852176053517136070371409409780081,
%U A295391 411179093017233901729922229390651516928263091167446329166300037456998815010841080003014976133796409791
%N A295391 a(1) = 2; a(n+1) = 1 + (a(n)^2 - a(n))*(1+1/n) for n >= 1.
%C A295391 (a(1), ..., a(n-1), a(n)-1) is an integer solution to the equation Sum_{k=1}^n k/x(k) = 1.
%H A295391 Robert Israel, <a href="/A295391/b295391.txt">Table of n, a(n) for n = 1..12</a>
%H A295391 R. Israel, <a href="https://math.stackexchange.com/questions/2531306/set-of-solutions-to-a-sum/2531415#2531415">Set of solutions to a sum</a>, Mathematics StackExchange.
%H A295391 N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=3RAYoaKMckM">A Nasty Surprise in a Sequence and Other OEIS Stories</a>, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; <a href="https://sites.math.rutgers.edu/~zeilberg/expmath/sloane85BD.pdf">Slides</a> [Mentions this sequence]
%F A295391 a(n) = A275611(n)/A275611(n-1). - _N. J. A. Sloane_, Sep 03 2024
%e A295391 1/1 = 1.
%e A295391 1/2 + 2/4 = 1.
%e A295391 1/2 + 2/5 + 3/30 = 1.
%e A295391 1/2 + 2/5 + 3/31 + 4/1240 = 1.
%e A295391 1/2 + 2/5 + 3/31 + 4/1241 + 5/1923550 = 1.
%p A295391 a[1]:= 2:
%p A295391 for n from 1 to 10 do a[n+1]:= 1 + (a[n]^2 - a[n])*(1+1/n) od:
%p A295391 seq(a[i],i=1..11);
%t A295391 Fold[Append[#1, 1 + (Last[#1]^2 - Last[#1]) (1 + 1/#2)] &, {2}, Range@ 8] (* _Michael De Vlieger_, Nov 21 2017 *)
%Y A295391 Closely related to A275611.
%K A295391 nonn
%O A295391 1,1
%A A295391 _Robert Israel_, Nov 21 2017