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# number game

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So your sequence is 27, 18, 12, 8, 4, 2, 1? It doesn't make much sense to just have a finite sequence of those seven numbers. For the sequence to be really general, you need to give us a way to generate additional terms.

It looks like the next term in your sequence would be (3 x 3 x 4) = 36, followed by (3 x 4 x 4) = 48. If this is correct, it seems that the first sequence in GuyZero's first link is correct. This is the sequence of maximal products of three numbers with the sum n. a(n) = max(r*s*t); n = r+s+t, offset by two (i.e. the first term in this sequence is the n = 3 term).

If additional terms are not as described above, you have a different sequence.

posted by mr_roboto at 12:18 PM on March 13, 2009

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# number game

March 12, 2009 12:09 PM Subscribe

Help me identify this arithmetic progression: (3 x 3 x 3) + (2 x 3 x 3) + (2 x 2 x 3) + (2 x 2 x 2) + (1 x 2 x 2) + (1 x 1 x 2) + (1 x 1 x 1)

does it have a name?

does it have a name?

those sequences expand to infinity, whereas mine goes from large to small. do you think it's really the same?

posted by Jason and Laszlo at 12:19 PM on March 12, 2009

posted by Jason and Laszlo at 12:19 PM on March 12, 2009

Uh, for some definition of "same".

I used (1x1x1), (1x1x2),(1x2x2),(2x2x2),(2x2x3), etc

You could also do (1x1x1),(1x1x1)+(1x1x2),(1x1x1)+(1x1x2)+(1x2x2), etc as a series I suppose.

What you posted technically isn't a series, it's just a sum. And since it goes backwards it's finite and there's not really much to say about finite integer sequences. Or sums. I mean, what you posted, taken literally, is 80. So I reversed it. And un-summed it. Otherwise it's just 80.

posted by GuyZero at 12:24 PM on March 12, 2009

I used (1x1x1), (1x1x2),(1x2x2),(2x2x2),(2x2x3), etc

You could also do (1x1x1),(1x1x1)+(1x1x2),(1x1x1)+(1x1x2)+(1x2x2), etc as a series I suppose.

What you posted technically isn't a series, it's just a sum. And since it goes backwards it's finite and there's not really much to say about finite integer sequences. Or sums. I mean, what you posted, taken literally, is 80. So I reversed it. And un-summed it. Otherwise it's just 80.

posted by GuyZero at 12:24 PM on March 12, 2009

72, not 80.

posted by DevilsAdvocate at 12:58 PM on March 12, 2009

posted by DevilsAdvocate at 12:58 PM on March 12, 2009

What's (F2)? Is it

F(2) = (1x1) + (1x2) + (2x2)

or

F(2) = (1x1x1) + (1x1x2) + (1x2x2) + (2x2x2)?

(and, similarly, F(4) = (4x4x4) + (4x4x3)... or F(4) = (4x4x4x4) + (4x4x4x3)....)

posted by milkrate at 1:55 PM on March 12, 2009

F(2) = (1x1) + (1x2) + (2x2)

or

F(2) = (1x1x1) + (1x1x2) + (1x2x2) + (2x2x2)?

(and, similarly, F(4) = (4x4x4) + (4x4x3)... or F(4) = (4x4x4x4) + (4x4x4x3)....)

posted by milkrate at 1:55 PM on March 12, 2009

Something like SUM (limits n to n

might be a more generalized way to write it (apologies for the atrociously inelegant notation).

posted by Emperor SnooKloze at 4:38 PM on March 12, 2009

_{1})= n^{n!}+ ... + n^{n1!}might be a more generalized way to write it (apologies for the atrociously inelegant notation).

posted by Emperor SnooKloze at 4:38 PM on March 12, 2009

You haven't written out a arithmetic progression there. You've written down a single sum. What does the next term in the sequence look like?

The best name for what you have written in your question is '72'.

posted by mr_roboto at 4:42 PM on March 12, 2009

The best name for what you have written in your question is '72'.

posted by mr_roboto at 4:42 PM on March 12, 2009

I obviously know nothing about math so excuse my poor use of the proper terms. Let's instead say that F1 is (3 x 3 x 3), F2 is (2 x 3 x 3), F3 is (2 x 2 x 3), and so on. I know it is nothing like the Fibonacci, but it has that same feel, that there's a systematic reduction (or if you reverse it, an increase).

posted by Jason and Laszlo at 7:20 PM on March 12, 2009

posted by Jason and Laszlo at 7:20 PM on March 12, 2009

Hi Jason, The terms of the sequence are known as the "Hexagonal Prism Numbers". If you sum them up, it's the "Partial Sum of the Hexagonal Prism Numbers".

You're also writing the terms down slightly incorrectly, you need to reverse the terms and group them like this:

F(1) = 1x1x1

F(2) = 2x2x2 + 2x2x1 + 2x1x1

F(3) = 3x3x3 + 3x3x2 + 3x2x2

F(4) = 4x4x4 + 4x4x3 + 4x3x3

...

Then you get

F(1) = 1

F(2) = 14

F(3) = 57

F(4) = 148

Refer to the OEIS Search Results for more information on how to express the terms of the series and how they have been used mathematically.

posted by onalark at 7:28 AM on March 13, 2009

You're also writing the terms down slightly incorrectly, you need to reverse the terms and group them like this:

F(1) = 1x1x1

F(2) = 2x2x2 + 2x2x1 + 2x1x1

F(3) = 3x3x3 + 3x3x2 + 3x2x2

F(4) = 4x4x4 + 4x4x3 + 4x3x3

...

Then you get

F(1) = 1

F(2) = 14

F(3) = 57

F(4) = 148

Refer to the OEIS Search Results for more information on how to express the terms of the series and how they have been used mathematically.

posted by onalark at 7:28 AM on March 13, 2009

*F1 is (3 x 3 x 3), F2 is (2 x 3 x 3), F3 is (2 x 2 x 3)*

So your sequence is 27, 18, 12, 8, 4, 2, 1? It doesn't make much sense to just have a finite sequence of those seven numbers. For the sequence to be really general, you need to give us a way to generate additional terms.

It looks like the next term in your sequence would be (3 x 3 x 4) = 36, followed by (3 x 4 x 4) = 48. If this is correct, it seems that the first sequence in GuyZero's first link is correct. This is the sequence of maximal products of three numbers with the sum n. a(n) = max(r*s*t); n = r+s+t, offset by two (i.e. the first term in this sequence is the n = 3 term).

If additional terms are not as described above, you have a different sequence.

posted by mr_roboto at 12:18 PM on March 13, 2009

This thread is closed to new comments.

posted by GuyZero at 12:16 PM on March 12, 2009