%I
%S 1,2,6,4,120,6,840,24,36,120,83160,12,1081080,840,120,16,294053760,36,
%T 5587021440,60,840,83160,128501493120,24,900,1081080,7560,2520,
%U 93163582512000,120,2888071057872000,10080,83160,294053760,840,36,106858629141264000
%N Start with n and find the LCM of n and A140635(n), and continue until you have a number where A140635(m) = m.
%e a(5) = 120 because:
%e A140635(5) = 2 and LCM of 5 and 2 is 10
%e A140635(10) = 6 and LCM of 10 and 6 is 30
%e A140635(30) = 24 and LCM of 30 and 24 is 120
%e A140635(120) = 120 so a(5) = 120.
%o (PARI)
%o s(n) = my(nd=numdiv(n)); for(k=1, n, if(numdiv(k) == nd, return(k))); \\ A140635
%o a(n) = my(m=n, t=s(n)); while(1, m=lcm(m, t); t=s(m); if(m==t, return(m))); \\ _Daniel Suteu_, Feb 25 2019
%Y Cf. A140635, A005179.
%K nonn
%O 1,2
%A _J. Lowell_, Feb 25 2019
%E More terms from _Rémy Sigrist_, Feb 25 2019
