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A192293
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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.
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6
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OFFSET
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1,1
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COMMENTS
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Like A019281 but using anti-divisors.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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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PROG
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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]
for n in range(1, 10**4):
if 3*n == sum(antidivisors(sum(antidivisors(n)))):
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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