A069561 Start of a run of n consecutive positive numbers divisible respectively by first n primes.

2, 2, 8, 158, 788, 788, 210998, 5316098, 34415168, 703693778, 194794490678, 5208806743928, 138782093170508, 5006786309605868, 253579251611336438, 12551374903381164638, 142908008812141343558, 77053322014980646906358

Offset: 1

Author

Amarnath Murthy, Mar 22 2002

Keywords

• nonn

Comments

It is evident that from a(3) onwards terms must be congruent to 8 mod p(3)#, where p(n)# is the n-th primorial (A002110). In fact the sequence for A069561(n) == k (mod p(n)#) for k: 2, 2, 8, 788, 788, 210988, etc. This follows from the Chinese Remainder Theorem.

Examples

			a(5) = 788 as 788, 789, 790, 791 and 792 are divisible by 2, 3, 5, 7, and 11 respectively.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..350

Crossrefs

Cf. A072562.

Programs

Mathematica

f[n_] := ChineseRemainder[-Range[0, n - 1], Prime[Range[n]]]; Array[f, 17, 2] (* Robert G. Wilson v, Jan 13 2012 *)
(* This code uses memoization in calculating the coeff for the primorial assoc'ed with a(n) value to generate a(n+1), producing 1000 terms in under one second (on a 2017 Costco Dell 64-bit Windows 10 machine)*)
q[1] =0; q[2] =0;
q[n_]:= (ModularInverse[Product[Prime[i], {i,1,n-1}], Prime[n]] * Mod[Prime[n]-n+1-g[n-1], Prime[n]])  // Mod[#, Prime[n]]&;
g[1] =2; g[2] =2;
g[r_] :=g[r]= g[r-1] + q[r] * Product[Prime[i], {i,1,r-1}];
Array[g, 1000]
(* Christopher Lamb, Oct 19 2021 *)

PARI

a(n)=lift(chinese(vector(max(n,2),k,Mod(1-k,prime(k))))) \\ Charles R Greathouse IV, Jun 20 2015

Formula

log a(n) << n log n. - Charles R Greathouse IV, Jun 20 2015

Extensions

More terms to a(15) from Sascha Kurz, Mar 23 2002

Edited and extended by Robert G. Wilson v, Aug 09 2002