Can Doubling Elements in a Sequence Ensure Divisibility by Their Position?

  • Context: Graduate 
  • Thread starter Thread starter Ben1587
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the problem of modifying elements in the natural sequence 1, 2, 3, ... to ensure that the sum of the first k terms is divisible by k. The scope includes theoretical exploration and mathematical reasoning.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant suggests replacing every element from 2 onward with 1, resulting in the sum of the first k terms being k.
  • Another participant proposes replacing all elements except the k-th with 0, setting the k-th element to k (or some integer a).
  • A different participant claims that doubling every number in the sequence works, as it leads to a sum of k(k+1), which is divisible by k.
  • One participant raises a question about the interpretation of "some" in the context of potentially meaning "a finite subset."

Areas of Agreement / Disagreement

Participants present multiple competing views on how to achieve the divisibility condition, with no consensus reached on a single solution.

Contextual Notes

Some proposed solutions involve specific modifications to the sequence, while others suggest broader interpretations of the problem, indicating potential limitations in the assumptions made.

Who May Find This Useful

Readers interested in mathematical problem-solving, particularly in number theory and sequence manipulation, may find this discussion relevant.

Ben1587
Messages
8
Reaction score
0
Replace some elements of the natural sequence 1, 2, 3, ... such that after this replacement the sum of first k terms is divisible by k.

Any solutions?
 
Mathematics news on Phys.org
Replace every element from 2 onward with 1. Then the sum of the first k terms will be k.
 
Replace all elements (except the k:th) with 0. Let the k:th element be k (or ak for some integer a ;)).
 
Ben1587 said:
Replace some elements of the natural sequence 1, 2, 3, ... such that after this replacement the sum of first k terms is divisible by k.

Any solutions?

What if by some he means 'a finite subset'?
 
Doubling every number works.

sum(1 to k)=k(k+1)/2 so multiplying this by 2 yields k(k+1) which is of course divisible by k.

Still, that is replacing all of the numbers, but I think it is on the road to a more pleasing solution.

Njorl
 

Similar threads

Undergrad Is this a proof of the Collatz Conjecture?
  • · Replies 8 ·
Replies
8
Views
3K
Graduate Is this Recursive Pattern in the Collatz Sequence Known?
  • · Replies 3 ·
Replies
3
Views
2K
MHB Why is $p_i + \frac{k}{p_i}$ divisible by $3$ and $8$?
  • · Replies 4 ·
Replies
4
Views
3K
MHB Is the Sum of n^2 Terms in an Arithmetic Sequence Limited to 1?
  • · Replies 2 ·
Replies
2
Views
2K
MHB Proving Divisibility by 6 Using Mathematical Induction
  • · Replies 7 ·
Replies
7
Views
4K
Undergrad Understanding permutations and combinations in a coin toss experiment
  • · Replies 6 ·
Replies
6
Views
2K
Graduate About a natural extension of Collatz conjecture
  • · Replies 3 ·
Replies
3
Views
2K
High School Divide by 4 and multiply by 15
  • · Replies 14 ·
Replies
14
Views
2K
High School Recursive sequences - notation question
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad Asymptotics for finding the successors in a Binary Tree
  • · Replies 3 ·
Replies
3
Views
2K