OFFSET
1,1
COMMENTS
Is this a duplicate of A098188? - R. J. Mathar, Oct 04 2018
Note that the first 4 terms and the next 4 terms form two sociable groups. But then the next 8 terms belong to two distinct sociable groups, whereas in A098188 the integers are grouped by cycle.
From Hartmut F. W. Hoft, Aug 23 2023: (Start)
This sequence is A098188 in ascending order.
Among the 19 4-cycles listed in the link by J. O. M. Pedersen only four of the 6 possible patterns of relative sizes of the numbers in a cycle are realized. (End)
LINKS
J. O. M. Pedersen, Known Unitary Sociable Numbers of order four [Via Internet Archive Wayback-Machine]
J. O. M. Pedersen, Order 4 cycles, 2007.
Eric Weisstein's World of Mathematics, Unitary Sociable Numbers
MATHEMATICA
f[n_] := f[n] = Module[{s = 0}, s = Total[Select[Divisors[n], GCD[#, n/#] == 1 &]]; Return[s - n]]; isok1[n_] := isok1[n] = Quiet[Check[f[n] == n, 0]]; isok2[n_] := isok2[n] = Quiet[Check[f[f[n]] == n, 0]]; isok4[n_] := isok4[n] = Quiet[Check[f[f[f[f[n]]]] == n, 0]]; isok[n_] := isok[n] = isok4[n] && Not[isok1[n]] && Not[isok2[n]]; Monitor[Position[Table[isok[n], {n, 1, 408408000}], True], n] (* Robert P. P. McKone, Aug 24 2023 *)
PROG
(PARI) f(n) = sumdiv(n, d, if(gcd(d, n/d)==1, d)) - n;
isok4(n) = iferr(f(f(f(f(n)))) == n, E, 0);
isok2(n) = iferr(f(f(n)) == n, E, 0);
isok1(n) = iferr(f(n) == n, E, 0);
isok(n) = isok4(n) && !isok1(n) && !isok2(n);
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Michel Marcus, Oct 01 2018
STATUS
approved