A247825 Numbers which are the difference between the sum of their bi-unitary divisors and the sum of their unitary divisors.

24, 240, 360

Offset: 1

Author

Paolo P. Lava, Sep 29 2014

Keywords

• nonn
• more
• bref

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

Examples

			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.
		

Links

• C. R. Wall, Bi-unitary perfect numbers, Proc. Am. Math. Soc. 33 (1) (1972) 39-42.
• Wikipedia, Unitary divisor

Crossrefs

Cf. A034448, A064591, A188999.

Programs

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);

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