A281571 -id:A281571 - cp's OEIS frontend

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

Derek Orr, Jun 27 2014

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]

			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.

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

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++)

A386985 Smallest k > 0 such that the base-n number formed by concatenating k, k - 1, ..., 2, 1 (each written in base n) is prime, or -1 if no such k exists for the given n.

Original entry on oeis.org

2, 2, 4, 2, 2, 373, 2, 2, 82, 2, 3

Offset: 2

Marco Ripà, Aug 11 2025

If a(13) != -1, then the corresponding prime must have more than 4178 decimal digits.
Sequence continues n=13..36: ?, 2, 2, 28, 362, 2, ?, 2, 2, 4, 2, 3, ?, 2, 5, 4, 2, 2, 37, 3, 2, 4, 2, 2.
The increasing-order analog begins 15, 2, ?. See A048436.

			a(3) = 2 since 21_(base-3) = 7_(base-10), which is prime.

Mathematics Stack Exchange, Concatenation of consecutive positive integers: Does a prime also exist in base 4 (as in bases 2, 3, 5, 6, 7, 8, 9,...)?.
Robert G. Wilson v, Letter to N. J. A. Sloane, Sep. 1992.

Cf. A000422, A048436, A281571.

f(n, b) = my(p=1, L=1); sum(k=1, n, k*p*=L+(k==L&&!L*=b)); \\ adapted from A000422
a(n) = my(k=1); while (!ispseudoprime(f(k, n)), k++); k; \\ Michel Marcus, Aug 16 2025

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

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