A192293 Let sigma*_m (n) be the result of applying the sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma*_m (n) = k*n; this sequence gives the (2,3)-anti-perfect numbers.
32, 98, 2524, 199282, 1336968
Offset: 1
Examples
sigma*(32)= 3+5+7+9+13+21=58; sigma*(58)= 3+4+5+9+13+23+39=96 and 3*32=96. sigma*(98)= 3+4+5+13+15+28+39+65=172; sigma*(172)= 3+5+7+8+15+23+49+69+115=294 and 3*98=294. sigma*(2524)= 3+7+8+9+11+17+27+33+49+51+99+103+153+187+297+459+561+721+1683=4478; sigma*(4478)= 3+4+5+9+13+15+45+53+169+199+597+689+995+1791+2985=7572 and 3*2524=7572.
Programs
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Maple
with(numtheory): P:= proc(n) local i,j,k,s,s1; for i from 3 to n do k:=0; j:=i; while j mod 2 <> 1 do k:=k+1; j:=j/2; od; s:=sigma(2*i+1)+sigma(2*i-1)+sigma(i/2^k)*2^(k+1)-6*i-2; k:=0; j:=s; while j mod 2 <> 1 do k:=k+1; j:=j/2; od; s1:=sigma(2*s+1)+sigma(2*s-1)+sigma(s/2^k)*2^(k+1)-6*s-2; if s1/i=3 then print(i); fi; od; end: P(10^9);
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Python
from sympy import divisors def antidivisors(n): return [2*d for d in divisors(n) if n > 2*d and n % (2*d)] + \ [d for d in divisors(2*n-1) if n > d >=2 and n % d] + \ [d for d in divisors(2*n+1) if n > d >=2 and n % d] A192293_list = [] for n in range(1,10**4): if 3*n == sum(antidivisors(sum(antidivisors(n)))): A192293_list.append(n) # Chai Wah Wu, Dec 02 2014
Extensions
a(4)-a(5) from Chai Wah Wu, Dec 01 2014
Comments