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August 24

Concatenated primes

Consider the sequence of numbers given by concatenating the first n integers in reverse order (1, 21, 321, 4321...). The first prime value in the sequence occurs when n = 82. I haven't found any more for n <= 500. Are there any more prime numbers in the sequence? Are there infinite primes in the sequence? 150.135.210.86 (talk) 17:24, 24 August 2015 (UTC)

I don't know - but the OEIS is a great resource for this kind of thing - see their page and refs on the sequence here [1]. SemanticMantis (talk) 17:52, 24 August 2015 (UTC)

Heuristically there should be an infinite number of primes in the sequence: We know that a(n) isn't divisible by 2 and 5, but apart from that it probably behaves like a random integer for primality testing purposes, so by the Prime number theorem the probability that a(n) is prime is approximately $\frac{2}{1}\cdot \frac{5}{4}\cdot \frac{1}{\ln (a(n))}$. Excluding the range you've already tested, the expected number of primes remaining in the sequence is $\sum _{n=501}^{\mathrm{\infty }}2.5/\ln (a(n))$. Given that $a(n)<10^{n({\log }_{10}(n)+1)}$ and that $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln (n)}$ diverges, the expected number of primes is infinite. To get an idea how far you'd have to search to have a 50% chance of finding the next example, you could try solving $\sum _{n=501}^{N}2.5/\ln (a(n))\ge 1$ for N with a computer. Egnau (talk) 22:42, 25 August 2015 (UTC)

oeis:A176024 says the next term is for n = 37765, found by Eric W. Weisstein in 2010. It has 177719 digits and is only a probable prime so far. PrimeHunter (talk) 01:09, 26 August 2015 (UTC)

The corresponding sequence base 2 contains primes for n = 2, 3, 4, 7 at least. --JBL (talk) 23:02, 26 August 2015 (UTC)

Up to $n=1000$ we have 2, 3, 4, 7, 11, 13, 25, 97, 110. Strangely, this sequence isn't in OEIS. -- Meni Rosenfeld (talk) 10:01, 28 August 2015 (UTC)