A229861 -id:A229861 - cp's OEIS frontend

A229860 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (2,k)-anti-perfect numbers.

3, 5, 7, 8, 14, 16, 32, 41, 56, 92, 98, 114, 167, 507, 543, 946, 2524, 3433, 5186, 5566, 6596, 6707, 6874, 8104, 9615, 15690, 17386, 27024, 84026, 87667, 167786, 199282, 390982, 1023971, 1077378, 1336968, 1529394, 2054435, 2276640, 2667584, 3098834, 3978336

Offset: 1

Paolo P. Lava, Oct 01 2013

nonn

Tested up to n = 10^6.

			Anti-divisors of 92 are 3, 5, 8, 37, 61. Their sum is 114.
Again, anti-divisors of 114 are 4, 12, 76. Their sum is 92 and 92 / 92 = 1.

Cf. A066272, A066417, A019278, A019292, A019293, A192293, A214842, A229861, A229862.

with(numtheory); P:=proc(q,h) local a,i,j,k,n;
for n from 3 to q do a:=n; for i from 1 to h do
k:=0; j:=a; while j mod 2 <> 1 do k:=k+1; j:=j/2; od;
a:=sigma(2*a+1)+sigma(2*a-1)+sigma(a/2^k)*2^(k+1)-6*a-2; od;
if type(a/n,integer) then print(n); fi; od; end: P(10^6,2);

Offset corrected and a(34)-a(42) from Donovan Johnson, Jan 09 2014

A229862 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (4,k)-anti-perfect numbers.

Original entry on oeis.org

5, 6, 7, 8, 14, 16, 41, 46, 56, 58, 64, 92, 96, 114, 946, 3307, 3325, 5186, 5566, 6596, 6874, 7982, 8104, 14621, 17386, 27024, 44217, 45970, 84026, 91282, 135592, 167786, 1077378, 1231058, 1529394, 2667584, 2873910, 3098834, 3978336, 4292594, 4921776, 27914146

Offset: 1

Paolo P. Lava, Oct 01 2013

nonn

Tested up to n = 10^6.

			Anti-divisors of 58 are 3, 4, 5, 9, 13, 23, 39. Their sum is 96.
The only anti-divisor of 96 is 64.
Again, anti-divisors of 64 are 3, 43. Their sum is 46. Finally, anti-divisors of 46 are 3, 4, 7, 13, 31. Their sum is 58 and 58 / 58 = 1.

Cf. A066272, A066417, A019278, A019292, A019293, A192293, A214842, A229860, A229861.

with(numtheory); P:=proc(q,h) local a,i,j,k,n;
for n from 5 to q do a:=n; for i from 1 to h do
k:=0; j:=a; while j mod 2 <> 1 do k:=k+1; j:=j/2; od;
a:=sigma(2*a+1)+sigma(2*a-1)+sigma(a/2^k)*2^(k+1)-6*a-2; od;
if type(a/n,integer) then print(n); fi; od; end: P(10^6,4);

Offset corrected and a(33)-a(42) from Donovan Johnson, Jan 09 2014

cp's OEIS Frontend

A229860 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (2,k)-anti-perfect numbers.

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A229862 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (4,k)-anti-perfect numbers.

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cp's OEIS Frontend

A229860 Let sigma*_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma*_m (n) = k*n; sequence gives the (2,k)-anti-perfect numbers.

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A229862 Let sigma*_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma*_m (n) = k*n; sequence gives the (4,k)-anti-perfect numbers.

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A229860 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (2,k)-anti-perfect numbers.

A229862 Let sigma_m (n) be result of applying sum of anti-divisors m times to n; call n (m,k)-anti-perfect if sigma_m (n) = k*n; sequence gives the (4,k)-anti-perfect numbers.