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07-18-09, 22:26
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Registered User
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Join Date: Jul 2009
Posts: 150
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The common solution on integer shown by author:
Quote:
with perfect (number, divider, sum_divider, show_d, limit) as ( select 2, 1, 1, varchar('1 ', 10000), 500
from sysibm.sysdummy1
union all
select number,
divider + 1,
case
when number = (divider + 1) * int(number / (divider + 1))
then sum_divider + (divider + 1)
else sum_divider
end,
case
when number = (divider + 1) * int(number / (divider + 1))
then show_d || ' + ' || varchar(divider + 1)
else show_d
end
,
limit
from perfect
where divider + 1 <= number / 2
and number <= limit
and divider + 1 <= limit / 2
and sum_divider <= number
union all
select number + 2, 1, 1, '1 ', limit
from perfect
where
( divider + 1 > number / 2
and number + 2 <= limit )
or
( divider + 1 <= number / 2
and number + 2 <= limit
and divider <= limit / 2
and sum_divider > number )
)
select number "perfect number", show_d
from perfect p1
where
p1.number =
(select max(p2.sum_divider) from perfect p2
where p2.number = p1.number)
and
p1.divider =
(select max(p2.divider) from perfect p2
where p2.number = p1.number)
;
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But it's take too long and needs some optimizitation.
Kara Sw., NY
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07-21-09, 12:51
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Registered User
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Join Date: Jul 2009
Location: NY
Posts: 886
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Why 1 is not a perfect number ?
1 = Sum(1). It could be only one known odd perfect number.
Lenny (Citigroup)
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07-21-09, 16:34
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Registered User
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Join Date: Sep 2004
Location: Belgium
Posts: 1,079
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Quote:
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Originally Posted by Lenny77
Why 1 is not a perfect number ?
1 = Sum(1).
It could be only one known odd perfect number.
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With that definition, 1 would be the only perfect number, since it's the only positive integer for which the sum of its divisors equals that number.
Since all other numbers n have 1 and itself as two divisors (at least), the sum of their divisors is always at least n+1.
(Actually, for a prime number n, it will be exactly n+1; for all other numbers, the sum will be strictly larger than n+1.)
For a perfect number n, Sum(divisors of n) = 2n.
__________________
--_Peter Vanroose,
__IBM Certified Database Administrator, DB2 9 for z/OS
__IBM Certified Application Developer
__ABIS Training and Consulting
__http://www.abis.be/
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07-21-09, 18:24
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Registered User
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Join Date: Jul 2009
Location: NY
Posts: 886
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How I know the problem with the odd perfect numbers did not solve yet.
Even nobody knows about exists or not exists such numbers.
It could be one of them.
Lenny
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09-29-09, 17:00
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Registered User
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Join Date: Jul 2009
Location: NY
Posts: 886
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New way to get small perfect numbers
The new way to get small perfect numbers:
Code:
with
Input(Lim) as
(select int(1e3) from sysibm.sysdummy1
)
,
Perfect_No (perf_cand, fact_sum, fact, lim) as
(select int(4), int(1), int(1), lim
from Input
union all
select perf_cand, fact_sum, fact + 1, lim
from Perfect_No
where mod(perf_cand, fact + 1) > 0
and fact + 1 <= sqrt(perf_cand)
and fact_sum <= perf_cand
union all
select perf_cand, fact_sum + (fact + 1) + (perf_cand / (fact + 1)), fact + 1, lim
from Perfect_No
where mod(perf_cand, fact + 1) = 0
and fact + 1 <= sqrt(perf_cand)
and fact_sum <= perf_cand
union all
select perf_cand + 1, 1, 1, lim
from Perfect_No
where (fact + 1 > sqrt(perf_cand) or fact_sum > perf_cand)
and perf_cand + 1 <= lim
)
,
Just_A_Perfect (perfect_number) as
(
select distinct perf_cand
from Perfect_No p1
where perf_cand = (select max(fact_sum)
from Perfect_No p2 where p1.perf_cand = p2.perf_cand )
)
select perfect_number from Just_A_Perfect
As Result:
Lenny
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The Perfect Number
