[seqfan] Re: Unitary RMPN

[hv at crypt.org]

Using a crude exhaustive search I can confirm:
a(6) = 216 [ 2^3 * 3^3 ]
a(10) = 5,292,000 [ 2^5 * 3^3 * 5^3 * 7^2 ]
a(12) = 10,584,000 [ 2^6 * 3^3 * 5^3 * 7^2 ]
a(14) > 784,000,000

Not sure how to make any further progres; can anyone prove even that
a(n) exists for all n?

Hugo

Earlier I wrote:
:Hi, each prime factor of n must divide a(n), and each prime power dividing
:a(n) must be >= n. That implies 72 divides a(6), and it is easy to show
:72 and 144 do not satisfy the requirement but 216 does, so a(n)=216.
:
:For n=10 we require 400 dividing a(10); working by hand, I found a better
:solution a(10) ?= 2^5 . 3^3 . 5^3 . 7^2; I don't know if that's the first,
:but it seems likely.
:
:I don't know of an effective way to find these for the general case though,
:or to verify a value is the first possible, other than by crude exhaustive
:search.
:
:Hugo
:
:zbi74583.boat at orange.zero.jp wrote:
::    Hi,Seqfan
::
::    [ Unitary Rational Multiply Perfect Number ]
::
::    Definition of URMPN :
::    UnitarySigma(n)=k*n
::              Where k is rational number
::    Ex.
::    k=3/2
::    2,20,24,360,816,....
::    k=5/3
::    12,18,2^6*3*7*13,2^6*3^2*5*7*13,2^15*3^6*5*7*11*19*37*73*83*331
::
::    Definition of ((n+1)/n )URMPN :
::    UnitarySigma(m)=(n+1)/n*m
::             Wnere n is positive integer
::
::    Definition of Sequence a(n) :
::    The smallest number which is ((n+1)/n)URMPN
::    6,2,3,4,5,216,7,8,9,2^15*3^3*5^4*7^2*79*83*157*313*331,11,-,13,-,-,16,17,-,19,....
::             Where "-" means unknown
::
::    If n is prime power then a(n)=n
::
::    Could anuone confirm a(6) and a(10) and compute more terms which are not
::prime power?
::    I am sure that a(12) has many digit
::
::
::
::    Yasutoshi
::
::
::
::
::
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