From Newsgroup: rec.puzzles

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

b) 1, 5, 14, 30, 55, ?

c) If these above sequences are allowed to continue indefinitely are there
any numbers, other than (zero and) one, common to both sequences? If not,
why not?

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...
--
David Entwistle
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From Newsgroup: rec.puzzles

David Entwistle used his keyboard to write :

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

starting from the last numer, 35 is the sum of previous four numbers,
so the next could be 69

b) 1, 5, 14, 30, 55, ?

and this would be 104

c) If these above sequences are allowed to continue indefinitely are there any numbers, other than (zero and) one, common to both sequences? If not, why not?

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...

--
/-\ /\/\ /\/\ /-\ /\/\ /\/\ /-\ T /-\
-=- -=- -=- -=- -=- -=- -=- -=- - -=-
........... [ al lavoro ] ...........
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From Newsgroup: rec.puzzles

In article <105ndpr$35ler$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

Yes

b) 1, 5, 14, 30, 55, ?

Yes

c) If these above sequences are allowed to continue indefinitely are there >any numbers, other than (zero and) one, common to both sequences? If not, >why not?

Not immediately obvious to me. If a(m) = b(n) then m/n is close to
the cube root of 2.

An easier question:

d) Give a geometric explanation for why b(n) = a(n) + a(n-1)

-- Richard
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 11:45:24 -0000 (UTC), Richard Tobin wrote:

d) Give a geometric explanation for why b(n) = a(n) + a(n-1)

SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.

This is the three-dimensional equivalent of the relationship between
square and triangular numbers. Working from the base of both tetrahedra,
we can arrange the two triangles into a rhombus and then stack the smaller layers on top to produce a rhombus-based pyramid. With a bit of distortion
we can imagine making it square-based.

When I was reading about square and triangular numbers, a few weeks ago, I though it was interesting. I didn't realize at the time that a triangular number IS a triangular number and IS NOT a square number and vice versa -
I think that is right. If you are a square number, you are not a
triangular number. Having looked at this a bit, I think that's true for a tetrahedral number and a square-based-pyramid number, but I can't prove it
as yet. Perhaps I should confirm that view is correct for triangular and square numbers first.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 10:19:49 +0200, Ammammata wrote:

starting from the last numer, 35 is the sum of previous four numbers,
so the next could be 69

b) 1, 5, 14, 30, 55, ?

and this would be 104

Interesting, but not what I was thinking of.
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <105o2mu$fvjs$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

When I was reading about square and triangular numbers, a few weeks ago, I >though it was interesting. I didn't realize at the time that a triangular >number IS a triangular number and IS NOT a square number and vice versa -
I think that is right. If you are a square number, you are not a
triangular number.

36

-- Richard
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From Newsgroup: rec.puzzles

In article <105o2mu$fvjs$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Having looked at this a bit, I think that's true for a
tetrahedral number and a square-based-pyramid number, but I can't prove it >as yet.

According to https://mathworld.wolfram.com/TetrahedralNumber.html

Beukers (1988) has studied the problem of finding numbers which are
simultaneously tetrahedral and pyramidal via integer points on an
elliptic curve, and finds that the only solution is the trivial
Te_1 = P_1 = 1.

But I can't find the paper online.

-- Richard
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 07:15:07 -0000 (UTC), David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Following Richard Tobin ....

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

Yes.

b) 1, 5, 14, 30, 55, ?

Yes.

c) If these above sequences are allowed to continue indefinitely are there >any numbers, other than (zero and) one, common to both sequences? If not, >why not?

If you need some a(n) to be equal to some b(m), you need to find the
integer solutions to

n(n+1)(n+2) = m(m+1)(2m+1)

Higher order Diophantine equations scare the hell out of me!

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...

So, I will lean to saying "No", but without proof. Perhaps, one
can find a proof by contradiction or something like that without
explicitly solving the above Diophantine equation. Have to think
about it.
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 11:23:02 -0400, Charlie Roberts wrote:

n(n+1)(n+2) = m(m+1)(2m+1)

Yes, that's where I got to. LHS is the product of three consecutive
integers. I'm wondering if there is something specific about factors...
--
David Entwistle
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 14:08:22 -0000 (UTC), Richard Tobin wrote:

36

Ah!
--
David Entwistle
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From Newsgroup: rec.puzzles

On 7/22/2025 12:15 AM, David Entwistle wrote:

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

b) 1, 5, 14, 30, 55, ?

c) If these above sequences are allowed to continue indefinitely are there any numbers, other than (zero and) one, common to both sequences? If not,
why not?

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...

When I see sequence puzzles like this, I first try building difference triangles. When the differences become the same, or match a previous
row, it is simple to go back up the triangle and find the next number.
In this case, for a) this is 56 and b) this is 91.

1 4 10 20 35 : 56
3 6 10 15 : 21
3 4 5 : 6
1 1 : 1

1 5 14 30 55 : 91
4 9 16 25 : 36
5 7 9 : 11
2 2 : 2

When the differences don't result in a row where all of the two or more numbers are the same, or match a previous row, then the sequence is
likely based on trivia*. Sequences based on trivia can be frustrating.
I have worked on some sequences where the pattern is based on a
localized fact (e.g., train stops of the London rail system), or outside
my interests (e.g., scores in hockey playoff games). It would helpful
if someone posting a sequence puzzle would state that the sequence is
based on mathematical manipulations or upon trivia (or both). If it is trivia, they should give clues (sports localized to Canada).
--
Carl G.


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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 15:26:06 -0000 (UTC), David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On Tue, 22 Jul 2025 11:23:02 -0400, Charlie Roberts wrote:

n(n+1)(n+2) = m(m+1)(2m+1)

Yes, that's where I got to. LHS is the product of three consecutive >integers. I'm wondering if there is something specific about factors...

The fact that the LHS is zero mod 6 seemed promising, but I
cannot make any progress.
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From Newsgroup: rec.puzzles

In article <105o6i4$1p358$2@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

Beukers (1988) has studied the problem of finding numbers which are
simultaneously tetrahedral and pyramidal via integer points on an
elliptic curve, and finds that the only solution is the trivial
Te_1 = P_1 = 1.

But I can't find the paper online.

This one?

"On oranges and integral points on certain plane cubic curves"

http://www.math.rug.nl/~top/oranges.pdf

Gareth
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From Newsgroup: rec.puzzles

In article <ttm*vj-hA@news.chiark.greenend.org.uk>,
Gareth Taylor <gtaylor@chiark.greenend.org.uk> wrote:

In article <105o6i4$1p358$2@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

Beukers (1988) has studied the problem of finding numbers which are
simultaneously tetrahedral and pyramidal via integer points on an
elliptic curve, and finds that the only solution is the trivial
Te_1 = P_1 = 1.

But I can't find the paper online.

This one?

"On oranges and integral points on certain plane cubic curves"

http://www.math.rug.nl/~top/oranges.pdf

That's the one.

-- Richard
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 18:37:07 -0000 (UTC), richard@cogsci.ed.ac.uk
(Richard Tobin) wrote:

In article <ttm*vj-hA@news.chiark.greenend.org.uk>,
Gareth Taylor <gtaylor@chiark.greenend.org.uk> wrote:

In article <105o6i4$1p358$2@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

Beukers (1988) has studied the problem of finding numbers which are
simultaneously tetrahedral and pyramidal via integer points on an
elliptic curve, and finds that the only solution is the trivial
Te_1 = P_1 = 1.

But I can't find the paper online.

This one?

"On oranges and integral points on certain plane cubic curves"

http://www.math.rug.nl/~top/oranges.pdf

That's the one.

-- Richard

Wow! The very equation!! And the solution.

Cubic curves are on the "To Do" list ... so
understanding this will have to wait.
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 18:37:07 -0000 (UTC), Richard Tobin wrote:

"On oranges and integral points on certain plane cubic curves"

http://www.math.rug.nl/~top/oranges.pdf

That's the one.

Amazing - I didn't realize it was such a difficult question. I hope posing
it didn't cause any anxiety.

Thanks for all the inputs.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Tue, 22 Jul 2025 09:15:36 -0700, Carl G. wrote:

It would helpful
if someone posting a sequence puzzle would state that the sequence is
based on mathematical manipulations or upon trivia (or both). If it is trivia, they should give clues (sports localized to Canada).

In this case, both sequences are mathematical manipulations and both
sequences are polynomial.
--
David Entwistle
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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

b) 1, 5, 14, 30, 55, ?

c) If these above sequences are allowed to continue indefinitely are there any numbers, other than (zero and) one, common to both sequences? If not, why not?

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...

Just so you know, yesterday, i started answering this on a yellow
sticky, and now I'm about fifteen sheet of paper deep, still
working. The next number on A and B were quite sinmple. I decided to
take a deep dive after realizing that I wasn't going to derive an N
expression anytime soon.

Part C is going to require a proof, and it's been twenty years since
graduating with a degree in math.

I'm having fun with it. I've discovered what the A sequence is, I will
reveal all when submitting my answer.

Working on the B sequence. Doing the matrices at the moment, it's been
forever since I've done Gauss-Jordan elimination. Had to pull my old
linear algebra textbook off the dusty shelf and do some review.

Give me a few days and I'll scan my notes and upload it somewhere so you
can see it.

I'm purposely avoiding the replies so I can do this on my own.

Thanks for this challenge. This sort of thing is why I chose to study
math in college years ago.

Daniel
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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

On Tue, 22 Jul 2025 14:08:22 -0000 (UTC), Richard Tobin wrote:

36

Ah!

More generally:
"""
A001110
Square triangular numbers: numbers that are both triangular and square. (Formerly M5259 N2291)
0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961,
113422539294030403250144100
"""
https://oeis.org/A001110

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

Can you add the next value to the following integer sequences a) and b)?

a) 1, 4, 10, 20, 35, ?

b) 1, 5, 14, 30, 55, ?

c) If these above sequences are allowed to continue indefinitely are there any numbers, other than (zero and) one, common to both sequences? If not, why not?

I'm okay with a) and b). I have some thoughts on c), but nothing
definitive. I think I may be becoming obsessed...

a: 56
b: 91
c: The answer is no. This is not a mundane question. I pulled out some
dusty college textbooks and did a ton of reading. I also contacted my
old academic advisor (retired some years ago) and discussed it a bit
with him.

The initial stack of sheets were complete messes. I did some rewriting
to clean it up and scanned my work. I wrote a proof for part C since
the question regards a sequence of positive integers to infinity and it
asked why. It sort of read like a number theory takehome quiz or
something from a discrete math class. It was not an easy question for
me.

Short summary:
Series A is described as x(x+1)(x+2)/6.
Series B is described as y(y+1)(2x+1)/6.

Both functions diverge almost instantly, no commonality between each.

I scanned the work and uploaded it in the following binary newsgroup:

ab.alt.binaries.misc

I looked for a binary ng that was relevant to rec.puzzles but came up
empty handed. So I looked for something generic and active. Posts are
named:

Math solution to number series p1
Math solution to number series p2

Part 1 is an image of both functions plotted on a cartesian plane and how quickly they diverge.

Part 2 is a scan of my work. I had to do some linear algebra to describe
each series, and haven't done Gauss-Jordan elimination in over twenty
years. It felt really good to do some fun math once again.

Thank you for the post. I had tons of fun relearning row reduced echelon
form. I messed up and restarted so many times before I got them
right. And my mechanical pencil got some heavy use. Went through at
least a dozen refill leads on this problem. It was a pleasure.

D
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From Newsgroup: rec.puzzles

On Thu, 31 Jul 2025 01:38:05 -0700, Daniel wrote:

Thank you for the post. I had tons of fun relearning row reduced echelon form. I messed up and restarted so many times before I got them right.
And my mechanical pencil got some heavy use. Went through at least a
dozen refill leads on this problem. It was a pleasure.

Phew! I'm glad you enjoyed it. I would have been mortified if I had caused
any distress. Well done too - quite an achievement well beyond my
abilities.

I don't know if you read all the posts, but Richard tobin pointed out that there is a paper on-line, concerning this problem, by Frits Buekers and
Jaap Top. They may be particularly interested in your approach.

https://www.math.rug.nl/~top/oranges.pdf

It looks like Buekers retired in 2019, but his email address and some of
his papers are still on-line.

https://webspace.science.uu.nl/~beuke106/

Top's contact details and work are also on-line.

https://www.math.rug.nl/~top/

The puzzle started out as a request to find the smallest number of cannon balls that can be arranged in either a triangular pyramid, or a square-
based pyramid, without any left over, but then I couldn't find a
solution...
--
David Entwistle
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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

On Thu, 31 Jul 2025 01:38:05 -0700, Daniel wrote:

Thank you for the post. I had tons of fun relearning row reduced echelon
form. I messed up and restarted so many times before I got them right.
And my mechanical pencil got some heavy use. Went through at least a
dozen refill leads on this problem. It was a pleasure.

Phew! I'm glad you enjoyed it. I would have been mortified if I had caused any distress. Well done too - quite an achievement well beyond my
abilities.

My degree is in applied mathematics and statistics. No, it was not an
easy degree. Yes, twenty years later means I'm no longer educated in
it.

I don't know if you read all the posts, but Richard tobin pointed out that there is a paper on-line, concerning this problem, by Frits Buekers and
Jaap Top. They may be particularly interested in your approach.

When I first opened the post, I saw the thread. After reading the
original post, I decided to solve it without seeing the responses.

Then, simply, forgot to check after posting late last night.

https://www.math.rug.nl/~top/oranges.pdf

Their approach using the sums and the k^2 identity was my initial
direction. In the end, both my approaches yielded the same result and
only posted one of them. Honestly, I would've included it too but
wanted to keep my pdf within the 1mb limit before splitting was
necessary. I was going to include ALL my work and scratch crap but the
pdf was too large.

Did you see my graph and proof?

It looks like Buekers retired in 2019, but his email address and some of
his papers are still on-line.

https://webspace.science.uu.nl/~beuke106/

Top's contact details and work are also on-line.

https://www.math.rug.nl/~top/

The puzzle started out as a request to find the smallest number of cannon balls that can be arranged in either a triangular pyramid, or a square-
based pyramid, without any left over, but then I couldn't find a
solution...

There is none. I'm not going to bother contacting either of those
scholars. My shitty little proof is NOT a sort of quality I'd want to
share to those outside this newsgroup, especially those who publish high quality papers like the one you posted.

Thanks for the direct link to the pdf, it's easy to wget.

D
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From Newsgroup: rec.puzzles

On Thu, 31 Jul 2025 21:47:47 -0700, Daniel wrote:

Did you see my graph and proof?

No, I don't have access to that newsgroup at the moment. Just non-binaries
(is that okay to say that nowadays?).
--
David Entwistle
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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

On Thu, 31 Jul 2025 21:47:47 -0700, Daniel wrote:

Did you see my graph and proof?

No, I don't have access to that newsgroup at the moment. Just non-binaries (is that okay to say that nowadays?).

Okay, well I can always email it to you if you wish. The email address
in the 'from' field won't go anywhere. But if you replace the 1's with
i's then it will.

You can say anything nowadays as far as I'm concerned.
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From Newsgroup: rec.puzzles

On 01/08/2025 08:48, David Entwistle wrote:

On Thu, 31 Jul 2025 21:47:47 -0700, Daniel wrote:

Did you see my graph and proof?

No, I don't have access to that newsgroup at the moment. Just non-binaries (is that okay to say that nowadays?).

When did they repeal free speech?
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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From Newsgroup: rec.puzzles

Richard Heathfield <rjh@cpax.org.uk> writes:

On 01/08/2025 08:48, David Entwistle wrote:

On Thu, 31 Jul 2025 21:47:47 -0700, Daniel wrote:

Did you see my graph and proof?

No, I don't have access to that newsgroup at the moment. Just
non-binaries
(is that okay to say that nowadays?).

When did they repeal free speech?

They didn't, is my point.

And my parents raised me to have a thick skin, take a joke, and not be
offended over things that don't effect me.

But then, my childhood was in the seventies and eighties.
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