OFFSET
1,1
COMMENTS
No further terms up to 10^8. Is there a relation with 6, 60 and 90, the 3 only bi-unitary perfects? - Michel Marcus, Oct 05 2014
a(4), if it exists, is larger than 10^11. - Giovanni Resta, Apr 15 2017
LINKS
C. R. Wall, Bi-unitary perfect numbers, Proc. Am. Math. Soc. 33 (1) (1972) 39-42.
Wikipedia, Unitary divisor
EXAMPLE
Divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Unitary divisors are 1, 3, 8, 24 and their sum is 36. Bi-unitary divisors are 1, 2, 3, 4, 6, 8, 12, 24 and their sum is 60. Then 60 - 36 = 24.
Divisors of 240 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240. Unitary divisors are 1, 3, 5, 15, 16, 48, 80, 240 and their sum is 408. Bi-unitary divisors are 1, 2, 3, 5, 6, 8, 10, 15, 16, 24, 30, 40, 48, 80, 120, 240 and their sum is 648. Then 648 - 408 = 240.
MAPLE
Q:=proc(n) local a, e, p, f; a:=1 ; for f in ifactors(n)[2] do e:=op(2, f); p:=op(1, f);
if type(e, odd) then a:=a*(p^(e+1)-1)/(p-1); else a:=a*((p^(e+1)-1)/(p-1)-p^(e/2)); fi; od: a ; end:
P:=proc(h) local a, b, k, n;
for n from 1 to h do a:=divisors(n); b:=0;
for k from 1 to nops(a) do if gcd(a[k], n/a[k])=1 then b:=b+a[k]; fi; od;
if Q(n)-b=n then print(n); fi; od; end: P(10^6);
PROG
(PARI) up(p, e) = p^e+1;
bup(p, e) = my(ret = (p^(e+1) - 1)/(p-1)); if ((e % 2) == 0, ret -= p^(e/2)); ret;
isok(n) = f = factor(n); n == (prod(k=1, #f~, bup(f[k, 1], f[k, 2])) - prod(k=1, #f~, up(f[k, 1], f[k, 2]))); \\ Michel Marcus, Oct 05 2014
CROSSREFS
KEYWORD
nonn,more,bref
AUTHOR
Paolo P. Lava, Sep 29 2014
STATUS
approved