Problem 2595. Polite numbers. Politeness.

Created by Jan Orwat

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367 Solutions
112 Solvers

Last Solution submitted on Oct 11, 2020

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Problem Comments

4 Comments

4 Comments

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Daniel Pereira on 27 Jan 2016

There are some flaws when checking the solutions:

I think 15 has 4 combinations of sum of consecutive INTEGERS (as stated in the problem):
15 = 7+8 = 4+5+6 = 1+2+3+4+5 = 0+1+2+3+4+5; % 0 is integer.

or 1025 (you say 5, I say 9):
1025 = 1022+1023 = sum(203:207) = sum(98:107) = sum(29:53) = sum(5:45) = sum(-4:45) = sum(-28:53) = sum(-97:107) = sum(-202:207); % negative numbers are integers too.

Jan Orwat on 28 Jan 2016

I can see the flaw in the description now, I've missed repeating "positive", Thanks. Btw considering your interpretation 15 has 7 combinations: sum(-14:15),sum(-6:8),sum(-3:6),sum(0:5),sum(1:5),sum(4:6),sum(7:8). In this way 1025 should have 11 and conversion between our interpretations is "yours=2*mine+1"; to any of mine solutions of the form m:n, you can add "-m+1:n", the last thing is to add "-input+1:input"

John D'Errico on 21 Aug 2016

An interesting problem, enough so that I chose to solve it in three essentially different ways. As always, there are various ways to solve any problem. The first two ways were essentially constructive, so counting the set of solutions for any N. The last used a formulaic approach.

Rafael S.T. Vieira on 11 Aug 2020

Politeness is an integer sequence defined at https://oeis.org/A069283.

Solution Comments

Solution 2056983

1 Comment

1 Comment

Asif Newaz on 13 Dec 2019

why it is showing--"While evaluating the solution, the server encountered an error caused by temporary unavailability of MATLAB Service'' ...

Solution 939609

1 Comment

1 Comment

John D'Errico on 21 Aug 2016

So, doing a little reading about polite numbers, one finds that the politeness divisors is related to the number of odd divisors.

Solution 939594

1 Comment

1 Comment

John D'Errico on 21 Aug 2016

vectorized & constructive, explicitly counting all solutions for even and odd k

Solution 939566

1 Comment

1 Comment

John D'Errico on 21 Aug 2016

Brute force, with a while loop. crude as hell, but it works for a first pass.

Solution 814993

1 Comment

1 Comment

Daniel Pereira on 27 Jan 2016

Whit this solution you can check the correct number of combinations that summing consecutive integers give the input number.

Solution 679419

2 Comments

2 Comments

Varoujan on 2 Jun 2015

I had to cheat to get the test 1 to pass. The test requires that if N=1, then politeness=0. According to other reputable sources, politeness of 1 is 1.

Jan Orwat on 28 Jan 2016

then why politeness of 15 isn't 4?

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number theory oeis

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Problem 2595. Polite numbers. Politeness.

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Problem 2595. Polite numbers. Politeness.

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