A069561 - OEIS

%I #31 Dec 14 2021 20:12:50

%S 2,2,8,158,788,788,210998,5316098,34415168,703693778,194794490678,

%T 5208806743928,138782093170508,5006786309605868,253579251611336438,

%U 12551374903381164638,142908008812141343558,77053322014980646906358

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

%C 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.

%H Charles R Greathouse IV, <a href="/A069561/b069561.txt">Table of n, a(n) for n = 1..350</a>

%F log a(n) << n log n. - _Charles R Greathouse IV_, Jun 20 2015

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

%t f[n_] := ChineseRemainder[-Range[0, n - 1], Prime[Range[n]]]; Array[f, 17, 2] (* _Robert G. Wilson v_, Jan 13 2012 *)

%t (* 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)*)

%t q[1] =0; q[2] =0;

%t 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]]&;

%t g[1] =2; g[2] =2;

%t g[r_] :=g[r]= g[r-1] + q[r] * Product[Prime[i], {i,1,r-1}];

%t Array[g, 1000]

%t (* _Christopher Lamb_, Oct 19 2021 *)

%o (PARI) a(n)=lift(chinese(vector(max(n,2),k,Mod(1-k,prime(k))))) \\ _Charles R Greathouse IV_, Jun 20 2015

%Y Cf. A072562.

%K nonn

%O 1,1

%A _Amarnath Murthy_, Mar 22 2002

%E More terms to a(15) from _Sascha Kurz_, Mar 23 2002

%E Edited and extended by _Robert G. Wilson v_, Aug 09 2002