A049300 - OEIS

%I #36 Oct 25 2023 05:34:01

%S 2,2,2,2,114,9440,217128,60044,20332472,417086648,74959204292,

%T 187219155594,79622514581574,14478292443584,6002108856728918,

%U 12288083384384462,5814429911995661690,14719192159220252523420

%N Smallest number starting a longest interval of consecutive integers, each of which is divisible by at least one of the first n primes.

%C The length of such longest interval of consecutive integers is given by A058989(n), which is the first maximal gaps A048670(n) minus 1 in the reduced residue system of consecutive primorial numbers.

%C Let j(m)=A048669(m) be the Jacobsthal function, i.e., the maximal distance between integers relatively prime to m. Let m=2*3*5*...*prime(n). Then a(n) is the least k>0 such that k,k+1,k+2,...,k+j(m)-2 are not coprime to m. Note that a(n) begins (or is inside) a large gap between primes. - _T. D. Noe_, Mar 29 2007

%H Brian Kehrig, <a href="/A049300/b049300.txt">Table of n, a(n) for n = 1..54</a> (terms 1..24 from Max Alekseyev).

%H Mario Ziller and John F. Morack, <a href="https://arxiv.org/abs/1611.03310">Algorithmic concepts for the computation of Jacobsthal's function</a>, arXiv:1611.03310 [math.NT], 2016-2017 (see ancillary file "remainders.txt").

%F a(n) = 1 + A128707(A002110(n)). - _T. D. Noe_, Mar 29 2007

%e Between 1 and 7, all 5 numbers (2,3,4,5,6) are divisible either by 2,3 or 5. Thus a(3)=2, the initial term. Between 113 and 127 the 13 consecutive integers are divisible by 2,5,2,3,2,7,2,11,2,3,2,5,2, each from {2,3,5,7,11}. Thus a(5)=114, the smallest with this property.

%Y Cf. A002110, A048670.

%K hard,nonn

%O 1,1

%A _Labos Elemer_

%E More terms from _T. D. Noe_, Mar 29 2007

%E a(11)-a(12) from _Donovan Johnson_, Oct 13 2009

%E a(13) from _Donovan Johnson_, Oct 20 2009

%E a(14) and beyond from _Max Alekseyev_, Nov 14 2009