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Numbers which are the difference between the sum of their bi-unitary divisors and the sum of their unitary divisors.
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%I #23 Oct 02 2018 08:18:08

%S 24,240,360

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

%C 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

%C a(4), if it exists, is larger than 10^11. - _Giovanni Resta_, Apr 15 2017

%H C. R. Wall, <a href="http://dx.doi.org/10.1090/S0002-9939-1972-0289403-9">Bi-unitary perfect numbers</a>, Proc. Am. Math. Soc. 33 (1) (1972) 39-42.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Unitary_divisor">Unitary divisor</a>

%e 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.

%e 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.

%p 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);

%p 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 P:=proc(h) local a,b,k,n;

%p for n from 1 to h do a:=divisors(n); b:=0;

%p for k from 1 to nops(a) do if gcd(a[k],n/a[k])=1 then b:=b+a[k]; fi; od;

%p if Q(n)-b=n then print(n); fi; od; end: P(10^6);

%o (PARI) up(p, e) = p^e+1;

%o bup(p, e) = my(ret = (p^(e+1) - 1)/(p-1)); if ((e % 2) == 0, ret -= p^(e/2)); ret;

%o 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

%Y Cf. A034448, A064591, A188999.

%K nonn,more,bref

%O 1,1

%A _Paolo P. Lava_, Sep 29 2014