A unitary divisor of a number n is a number d such that d|n and gcd(d, n/d)=1. For example, 3 is a unitary divisor of 12 because gcd(3, 12/3)=gcd(3, 4)=1.
A unitary perfect number is a number that is the sum of its unitary divisors less than itself. For example, 60 is a unitary perfect number because its unitary divisors less than itself are 1, 3, 4, 5, 12, 15, and 20, and 1+3+4+5+12+15+20=60. Find all unitary perfect numbers less than 1000000.
(In reply to
re: just look up - spoiler by Steve Herman)
1.
Nothing wrong with the puzzle. I gave it my TU.
2.
Nothing wrong with it being extracted from Wikipedia,
OEIS or any math. textbook.
3.
Nothing wrong with not sharing my opinion about its
D-level. It was voiced prior to the publication.
4.
I just claim that the
effort needed to get it answered (i.e. the mental process through which one
reaches two numbers i.e. 90 and 87360) is significantly less
than recently defined on the Forum as fitting D3.
5.
There is no analytical
solution, the existence of the 5th UP number
is presently an unresolved math problem (both facts obscure to an innocent solver),
therefore
one is driven either to waste his time by trying in vain to "invent
the wheel" or to write a program (with a limit set by someone who knows how futile is a
longer trial), while both the 4 known numbers and a 2-line Mathematica
statement to generate them exist on OEIS.
The above facts caused me to add a discouraging comment re the current "state of art".
6.
I am neither the first nor
the only one to address the D-level issue.
Please read attentively "Two happy ends" (published in 2011), both
the the puzzle and the comments, that were
triggered by 1st solver remark.
The discussion that followed might be a
good example of professional dialog, fully resolving flaws and misjudgments and
sharpening the solving tools of all
involved.
<o:p> 7. Steve, the problem is not trivial. But one should not ignore the existence of OEIS and Internet, the best handbooks so easily accessable.
Anyway, thanks for sharing with me your criteria.</o:p>
re(2): just look up by SH
