#find a set of unique rationals in the form 1 over x that sum to 1 (with 2 <= x <= 2023)

79 messages · Page 1 of 1 (latest)

worn pawn
#

I need a formula or something to get the most unique rationals that up to 1 for a school challenge. Any help would be appreciated.

jade lanternBOT
#
  1. Wait patiently for a helper to come along.
  2. Once someone helps you, say thank you and close the thread with: ```diff
    +close
worn pawn
#

find a set of unique rationals in the form 1 over x that sum to 1 (with 2 <= x <= 2023)

dire ferry
#

@worn pawnhow many rationals are you using though?

pine yacht
worn pawn
worn pawn
pine yacht
worn pawn
velvet scroll
#

I am not sure I understand the question properly but you can use a recursive method

#

for all x>0 you have 1/x = 1/(x+1) + 1/(x(x+1))

#

so if you have a decomposition of 1 you can double the number of terms each time

#

1 = 1/2 + 1/3 + 1/6
= (1/3 + 1/6) + (1/4 + 1/12) + (1/7 + 1/42)

#

and so on

#

1 = 1/3 + 1/4 + 1/6 + 1/7 + 1/12 + 1/42
= (1/4 + 1/12) + (1/5 + 1/20) + (1/7 + 1/42) + (1/8 + 1/56) + (1/13 + 1/156) + (1/43 + 1/1806)

#

if you want decompositions with denominators <2023 you need to stop decomposing terms that have a denominator >44 (because 45*46 > 2023)

dire ferry
#

there called farey sequences

worn pawn
#

i got to 5q

#

52

#

and then i couldnt split them anymore

worn pawn
#

if you have alot of rationals

#

i have 942 rationals

#

thats the list of 942

#

it starts from 237

velvet scroll
#

i am not entirely sure what you want to do

#

if your point was just finding the biggest sum with bounded denominators you have pretty much stretched the process to the limit

#

I don't know if it is optimal but that's already plenty

#

if you wish you could keep going indefinitely with bigger denominators

worn pawn
#

if you put each of those numbers over 1

#

so 1/x

velvet scroll
#

yeah i know

worn pawn
#

and then add them all together

#

it equals to 1

velvet scroll
#

i understand

#

lol

worn pawn
#

so how couldi find more, its bugging me

velvet scroll
#

that wasn't my question

worn pawn
#

oh

#

so any ideas

velvet scroll
#

what are you trying to accomplish exactly

worn pawn
#

there are definitely more

worn pawn
velvet scroll
#

find the biggest list with bounded x <= 2023

velvet scroll
worn pawn
#

yea 2 <= x <= 2023

#

thats what x must be

#

in 1/x

velvet scroll
#

then it's an optimization problem

worn pawn
#

do you know howi could tackle it

velvet scroll
#

not really no

worn pawn
#

do you know anyone who could help me

velvet scroll
#

not really no

worn pawn
#

damn

velvet scroll
#

what are you going to do with your list ?

worn pawn
#

its for a competition

merry schooner
#

Axiom 1: π is rational and has numerator 1
Axiom 2: 1-π has numerator 1
Solution: π-π+1=1
π is very unique

pseudo temple
#

Well let's try to face it with some nt approach .

#

Sum of 1/n_i is the sum over all not n_i numbers by the multiplication of all ni's

#

But they both should be the same

#

Cuz..m it's 1

#

well if we take modn_i of the ewuatiob

#

We get:

#

Multiplication of all numberd which are not n_i in n is equal to 0 modn.

#

n1n2n3...ni-1*ni+1...nk..=modni to 0

#

Meaning
...

#

We need to find the largest sequence where this holds

#

For every number in the sequence

#

Well actually

#

My whole approach was stupid

#

Since mod adds more solutions...

#

My bad...

worn pawn
#

Damn

#

I really need something