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

1,1

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.

Links

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

Formula

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

Example

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

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 *)

Prog

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

Crossrefs

Cf. A072562.

Sequence in context: A270405 A047692 A270316 * A180370 A326939 A341303

Adjacent sequences: A069558 A069559 A069560 * A069562 A069563 A069564

Keyword

nonn

Author

Amarnath Murthy, Mar 22 2002

Extensions

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

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

Status

approved