A244424 Least number k > 0 such that concatenating k consecutive natural numbers beginning with n is prime, or 0 if no such number exists.

0, 1, 1, 4, 1, 2, 1, 2, 179, 0, 1, 2, 1, 4, 5, 28, 1, 3590, 1, 4, 0, 0, 1, 0, 25, 122, 0, 46, 1, 0, 1, 0, 71, 4, 569, 2, 1, 20, 5, 0, 1, 2, 1, 8, 0, 0, 1, 0, 193, 2, 0, 0, 1, 0, 0, 2, 5, 4, 1, 0, 1, 2, 0, 4, 5, 938, 1, 2, 119, 58, 1, 116, 1, 0, 125, 346, 5, 2, 1, 2, 0, 0, 1, 0, 0, 32

Offset

1,4

Comments

The current zero values are only conjectural: a(n) > 5000 - n for all a(n) = 0 shown. [Edited by M. F. Hasler, Apr 27 2017]

A positive value for a(1) will satisfy A075019(a(1)) = A007908(a(1)). - Michel Marcus, Jul 09 2014

Probably a(n) > 0 for all n. Appending k integers gives a number of size ~10^(k log_10 k) and so the expected number of primes with k < x is about the integral of 1/(k log k) up to x which is log log x. This diverges, so by the Borel-Cantelli lemma we expect that there will be a prime eventually. (Corrections for the particular base at hand affect the expected number but not its order of growth.) On the other hand, log log x grows slowly so finding the values of a(1), a(10), a(21), etc. may be hard. - Charles R Greathouse IV, Jul 10 2014 [Corrected by Pontus von Brömssen, Oct 12 2021]

Links

Table of n, a(n) for n=1..86.

Example

14 is not prime. 1415 is not prime. 141516 is not prime. 14151617 is prime. Thus a(14) = 4 since 4 consecutive numbers were concatenated.

Prog

(PARI) a(n) = {p=""; tot=0; for(i=n, 5000, p=concat(p, Str(i)); tot++; if(ispseudoprime(eval(p)), return(tot)))}
n=1; while(n<100, print1(a(n), ", "); n++)

Crossrefs

Cf. A007908, A075019, A281571 (base-2 variant).

Sequence in context: A256252 A247004 A010640 * A322574 A030787 A176218

Adjacent sequences: A244421 A244422 A244423 * A244425 A244426 A244427

Keyword

nonn,base,more,hard

Author

Derek Orr, Jun 27 2014

Status

approved