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If I have an infinite sequence of positive integers with infinite number of primes and if I have an infinite number of distinct sequences with such properties may I claim that there is an infinite number of primes on $n$ -th position,where $n$ is arbitrary number ?

Maybe this question is somehow related with probability but I don't see that relation.

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An interesting claim to investigate would be that there exists a number $n$ such that there is an infinite number of primes on the $n$th position. –  Sam Hocevar Nov 4 '11 at 23:28
    
@Sam Hocevar No, in a very similar method to my answer below, say you give me a collection of sequences that does satisfy the condition pedja stated or yours. Now, I add a 0 to the start of the first, two 0s to the start of the second, and on and on, n 0s to the start of the nth sequence. Then, there are only finitely many sequences that are NOT 0 in the nth position. So, clearly, there can not be infinitely many that are prime in the nth position. –  Graphth Nov 5 '11 at 17:00

4 Answers 4

up vote 8 down vote accepted

No. Suppose the first sequence is just the sequence of all primes: $$ 2,3,5,7,11,13,17,19,23,29,\ldots $$ and the second is the same except that $4$ appears where $2$ appeared before: $$ 4,3,5,7,11,13,17,19,23,29,\ldots $$ and the third sequence is the same except that $6$ appears where $2$ appeared in the first one: $$ 6,3,5,7,11,13,17,19,23,29,\ldots $$ the the fourth one is the same except that $8$ appears in the first position, and so on.

Then it is not true that there are infinitely many primes in the first position.

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No, because I could take any collection of sequences you gave me and I could add $n$ zeros to the beginning of each and I'd still have an infinite number of distinct sequences with an infinite number of primes each. Yet, there would be 0 primes in the 1st through $n$th position.

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what if all numbers in sequence are distinct ? –  pedja Nov 4 '11 at 19:21
    
@pedja That is trickier... The problem is, any sequence could possibly contain every positive integer, or every nonzero integer, or every integer. So, if I wanted to add in one at the beginning, or several at the beginning, there would literally be none left to add. But, in such a case, I can still exhibit a counterexample. Give me any such sequence that satisfies your condition but where all integers are used. For the $n$th sequence, I find the first $n$ powers of 2 (with power at least 2) in the sequence and move them to the front, creating new sequences that don't satisfy your condition –  Graphth Nov 5 '11 at 17:04

That conclusion doesn't follow from what you're given. For example, let the $k$th sequence be $\{2k+1,2k+2,2k+3,\dots\}$ (just all the integers in a row, starting from $2k+1$). Then each sequence contains infinitely many primes, but none of the sequences has a prime in the $4$th position (or indeed, in the $n$th position for any even number $n\ge4$).

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If I understand what you are saying, the answer would be "no".

For any subset $S$ of $\mathbb{N}$ that has infinite complement, there are uncountably many sequences of natural numbers that have no primes at all in any position corresponding to $S$ and have an infinite number of primes in positions corresponding to the complement. In particular, specify any collection of positions that has infinite complement, and there are uncountably many sequences, each of which has infinitely many prime terms, but that have no primes at all in the specified collection of positions.

You can even take an increasing sequence of subsets $S$ so that no specific position will have more than finitely many primes in them: start as Michael Hardy's answer. Then replace the first entry with a nonprime. Then replace the second entry with a nonprime. Then replace the third entry with a nonprime. Etc. Then position $n$ will have primes only in the first $n$ sequences of your list, so no position has primes infinitely often.

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