A049300 Smallest number starting a longest interval of consecutive integers, each of which is divisible by at least one of the first n primes.

2, 2, 2, 2, 114, 9440, 217128, 60044, 20332472, 417086648, 74959204292, 187219155594, 79622514581574, 14478292443584, 6002108856728918, 12288083384384462, 5814429911995661690, 14719192159220252523420

Offset: 1

Author

Labos Elemer

Keywords

• hard
• nonn

Comments

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.

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

Examples

			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.
		

Links

• Brian Kehrig, Table of n, a(n) for n = 1..54 (terms 1..24 from Max Alekseyev).
• Mario Ziller and John F. Morack, Algorithmic concepts for the computation of Jacobsthal's function, arXiv:1611.03310 [math.NT], 2016-2017 (see ancillary file "remainders.txt").

Crossrefs

Cf. A002110, A048670.

Formula

a(n) = 1 + A128707(A002110(n)). - T. D. Noe, Mar 29 2007

Extensions

More terms from T. D. Noe, Mar 29 2007

a(11)-a(12) from Donovan Johnson, Oct 13 2009

a(13) from Donovan Johnson, Oct 20 2009

a(14) and beyond from Max Alekseyev, Nov 14 2009