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  • Does a sequence of positive integers $a_n$ such that the sequence $a_n, \space a_n+1,\space a_n+2,\space \cdots, \space a_n+9$ contains exactly $5$ primes exist?
  • Is such sequence finite or infinite?
  • Does it have a general term? If it does, then kindly prove it and provide its limit as $n \to + \infty$.

$a_0 = 2$ because the sequence $2, \space 3,\space \cdots, \space 11$ contains exactly $5$ primes $2,\space 3,\space 5,\space 7,\space 11$

I found a finite list of prime and I couldn't find any other integer and I was wondering if such integers except from 2 exist.

  • What about a sequence for exactly $4$ primes?

3 Answers 3

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Of 10 consecutive numbers, exactly five are even, one or two are multiples of six and three or four ar emultiples of three. Thus at least one of the multiples of three is odd. This leaves at most four numbers that are neither divisible by two nor three. Therefore, unless $2$ and $3$ appear themselves among the primes, no five primes can occur.

For the problem with four primes, there are seemingly many solutions, e.g. if we require additionally that the first of the ten numbers is prime, I find 3, 5, 11, 101, 191, 821, 1481, 1871, 2081, 3251, 3461, 5651, 9431, 13001, 15641, 15731, 16061, 18041, 18911, 19421, 21011, 22271, 25301, 31721, 34841, 43781, 51341, 55331, 62981, 67211, 69491, 72221, 77261, 79691, 81041, 82721, 88811, 97841, 99131, 101111, 109841, 116531, 119291, 122201, 135461, 144161, 157271, 165701, 166841, 171161, 187631, 194861, 195731, 201491, 201821, 217361, 225341, 240041, 243701, 247601, 247991, 257861, 260411, 266681, 268811, 276041, 284741, 285281, 294311, 295871, 299471, 300491, 301991, 326141, 334421, 340931, 346391, 347981, 354251, 358901, 361211, 375251, 388691, 389561, 392261, 394811, 397541, 397751, 402131, 402761, 412031, 419051, 420851, 427241, 442571, 444341, 452531, 463451, 465161, 467471, 470081, 477011, 490571, 495611 when looking only up to 500000.

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I think that's it for 5 primes. A sequence of 10 consecutive positive integers has 5 even numbers. It also contains 3 multiples of 3, at least one of which is odd, leaving you with 4 odd numbers not divisible by 3. For $a_0$ through $a_0+9$, that's enough since it actually has primes divisible by 2 and 3, namely 2 and 3 themselves.

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Sequence of 5 primes in the set a0 to a11 can occur only once. i.e., 2,3,5,7,11 Reasons: for any other set of positive integers,

  1. Set will have 5 unique even numbers which will be divisible by 2.
  2. Set will have at least 3 numbers which will be multiples of 3.
  3. The above two sets combined will have at least 6 elements.

and hence it proves that there can't be 5 primes in the specified set!