probabilistic number theory

[wildrabbitt]

The ith partial sums of : 1, 3, 6, 10, 15, 21

are

for i = 1 the sum is 1

for i = 2 the sum is 1 + 3 = 4

for i = 3 the sum is 1 + 3 + 6 = 10

in fact, I might have known that the ith partial sum is the (i + 1)th triangular number.

/* EDIT : that's wrong actually : it's the ith tetrahedral number. Sorry about that.*/


There is a sequence of positive integers which contains no arithmetic sequences of length more than 2, whose partial sums
are less than any other sequence for which the same rule holds.

It begins,

1, 2, 4, 5, 10, 11, 13, 14, 28, 29, 31, 32, ......

[wildrabbitt]

Thanks for taking the time to read my maths by the way :)

[R.D. Silverman]

Originally Posted by LaurV

I don't exactly grasp that condition with partial sums (it may be because I am silly, or because I don't exactly know the English terminology). For example, does the sequence of triangular numbers satisfy the condition? 1, 3, 6, 10, 15, 21, .... (I think this has no arithmetic progression in it)
If not, why not?
If yes, your lemma is false...

Did you bother to check your assertion???? Clearly not, e.g. 1,28,55

[wildrabbitt]

You misunderstand me Silverman. I didn't mean or say the nth partial sums of the reciprocals.

I said the nth partial sums of the sequence of the positive integers with the condition stated :


1, 2, 4, 5, 10, 11, 13, 14, 28, 29, 31, 32, ......

So, the third partial sum of this sequence is 7

The third partial sum of

the sequence you gave

1,28,55

is clearly 84 and 7 < 84.

So you want to hear more?

[LaurV]

No, RDS was right, he was talking about my assertion that the triangular sequence has no arithmetic progression, and provided a counterexample. Indeed I didn't check it, and I say sorry, it was "feeling", but you would not need to go so high, because the lemma would be still false for the sequence where 55 is substituted to 56

Back to your sequence, now I understand what you mean, and I know how to construct the sequence you gave, such as the partial sums are minimal. This is ok up to now. We are fine here. From my side, it doesn't need a demonstration.

[wildrabbitt]

No he wasn't Laurv.

I think I've come to wrong place to show my proof.

You're all more concerned about trying to find a mistake by misinterpreting what I'm saying.

I clearly said :

The sequence of natural numbers which contains no arithmetic sequences of length more than 2, whose nth partial sums
are less than any other sequence which also contains no arithmetic sequences of length more than 2.



Silverman thought I said :

The reciprocals of natural numbers in the sequence of natural numbers which contains no arithmetic sequences of length more than 2, whose nth partial sums
are less than any other sequence which also contains no arithmetic sequences of length more than 2.



...which I didn't.


Laurv didn't understand what I wrote either.

There's nothing ambiguous about what I wrote. It's too simple for post-graduates.

[LaurV]

Man, stop that. We are all too concerned with RDS because he is usually pointing out mistakes in his own way. This time he didn't talk to you at all, he just pointed out that the 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, .... contains a progression with length 3, i.e. 1, 28, 55, contrary to my assumption that "I think it has no progression". That is all. Nobody has anything against you, and I am clear what you said, after you provided the example. My problem is English, not math. At least, not at this level. Please forgive the last posts and resume from post #30. You can post the demo if you want, but for me is clear now, and not necessary.

[wildrabbitt]

Sorry. I didn't read all of your post or you editted. I'll post the next bit soon.

[R.D. Silverman]

Originally Posted by LaurV

No, RDS was right, he was talking about my assertion that the triangular sequence has no arithmetic progression, and provided a counterexample. Indeed I didn't check it, and I say sorry, it was "feeling", but you would not need to go so high, because the lemma would be still false for the sequence where 55 is substituted to 56

The triangular number sequence has infinitely many 3-term progressions (as do the squares).
Hint: Pellian

n(n+1)/2 - 1 = k(k+1)/2 - n(n+1)/2; complete the square........

[R.D. Silverman]

Originally Posted by LaurV

My problem is English, not math. At least, not at this level. .

Actually, for a non-native speaker, your English is SUPERB.

[wildrabbitt]

Well, this lemma is totally crucial for my proof so I'm very carefully trying to prove it.

Here's where I'm at so far. I'm posting now because I know I'm taking my time. Please go easy on the crankometer. I will finish it unless one of you beats me to it.


http://www.mersenneforum.org/attachm...1&d=1442407534


http://www.mersenneforum.org/attachm...1&d=1442407663


I can understand that I'm probably putting you all off with my childish tantrums but I do appreciate yor patience.