

A244424


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


4



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
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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 A244424(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 k log k up to x which is log log x. This diverges, so by the BorelCantelli 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


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. A281571 for the base2 variant.
Sequence in context: A256252 A247004 A010640 * A322574 A087230 A030787
Adjacent sequences: A244421 A244422 A244423 * A244425 A244426 A244427


KEYWORD

nonn,base,more,hard


AUTHOR

Derek Orr, Jun 27 2014


STATUS

approved



