Arithmetic Sequence that doesn't contain any powers

apple2357

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I am trying to generate an infinite arithmetic sequence that does contain any powers ( other than 1).
For example, i know 7, 17, 27, 37, ... etc does not contain square numbers, because no square number can end in a 7.
And i could probably do something similar for cube numbers...

But, i am stuck trying to generate one that doesn't contain any powers ( squares, cubes etc.)
Just exploring a problem that occurred to me, not even sure if it is possible?
Any ideas of where to start would be helpful!
 
I am trying to generate an infinite arithmetic sequence that does contain any powers ( other than 1).
For example, i know 7, 17, 27, 37, ... etc does not contain square numbers, because no square number can end in a 7.
And i could probably do something similar for cube numbers...

But, i am stuck trying to generate one that doesn't contain any powers ( squares, cubes etc.)
Just exploring a problem that occurred to me, not even sure if it is possible?
Any ideas of where to start would be helpful!

One obvious such sequence is the sequence of prime numbers.
 
...i am stuck trying to generate one that doesn't contain any powers ( squares, cubes etc.)
2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 60 64 : whoa! 8^2 = 64
So stop at 60....or are you looking for an infinite one? If so, WHY?
 
2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 60 64 : whoa! 8^2 = 64
So stop at 60....or are you looking for an infinite one? If so, WHY?

Yeah infinite one... more challenging!

I wondering if there is something in exploring even and odd numbers and powers?
 
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Nope, but my son is. And he keeps bringing me problems!

I think you said that this problem is one you made up, not from your son's class. Am I right?

My guess would be that it is impossible, though I haven't tried proving that.
 
Ya...your problem is in many ways equivalent to:
digging a hole without hitting any pebbles or stones :rolleyes:
 
I think you said that this problem is one you made up, not from your son's class. Am I right?

My guess would be that it is impossible, though I haven't tried proving that.

Yeah, we discussed an extension. The original problem was generating a sequence without square numbers.
I have been playing on a spreadsheet with 4n, 4n+1, 4n+2, 4n+3 .. no luck yet though
 
I have been playing on a spreadsheet with 4n, 4n+1, 4n+2, 4n+3 .. no luck yet though
BUT...no matter what n's value is, that's a series of consecutive integers:
n=1: 4,5,6,7,...
n=9: 36,37,38,39....
May I suggest that you're completely losing your time :razz:
Plus something tells me that you have a ficticious son:confused:
 
He is not fictitious!

I meant the sequence 4n+2 doesn't contain any powers! lik 2, 6, 10, 14, 18, ....
Just messing on a spreadsheet, looking for a counter example!

Or 10, 110, 210, 310... i don't think any powers end in zero that don't have another zero preceding it?
 
He is not fictitious!

I meant the sequence 4n+2 doesn't contain any powers! lik 2, 6, 10, 14, 18, ....
Just messing on a spreadsheet, looking for a counter example!

Or 10, 110, 210, 310... i don't think any powers end in zero that don't have another zero preceding it?
I am truly not sure what you are even trying to do.

But the unit digit of the square of one of the decimal digits is in the set {0, 1, 4, 5, 6, 9}.

\(\displaystyle n = 10k + d \implies n^2 = 100k^2 + 20k + d^2.\)

Therefore any arithmetic sequence with a starting number that has 2, 3, 7, or 8 as its units digit and with an increment that is a multiple of 10 will contain no perfect squares.
 
Ok, i think i have some ideas here..

Take any even number ( i.e. of the form 2k) and raise it to any power, i.e. (2k)^m . All these powers will always be divisible by 4, so any arithmetic sequence that contains numbers which are divisible by 4 cannot guarantee to not include a power.

Take any odd number ( i.e. of the form 2k+1) and raise it to any power. i.e. (2k+1)^m. And here i am getting stuck..
 
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