A086447 a(n) = the least k such that prime(n+1)+prime(n+2)+...+prime(n+k) is a multiple of prime(n).

2, 2, 6, 6, 6, 6, 4, 8, 4, 30, 7, 31, 37, 67, 22, 60, 46, 38, 69, 13, 65, 76, 19, 163, 9, 52, 100, 84, 66, 136, 66, 119, 33, 79, 47, 76, 187, 214, 37, 96, 461, 111, 62, 189, 510, 37, 256, 121, 130, 132, 144, 481, 64, 195, 53, 47, 136, 90, 194, 318, 526, 151, 788, 1542

Offset: 1

Author

Zak Seidov, Jul 20 2003

Keywords

• easy
• nonn

Comments

Conjecture: a(n) exists for every n.

Examples

			a(3)=6 because prime(3)=5 divides 7+11+13+17+19+23 = 90.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A055233, A055514, A086448.

Programs

Mathematica

bb={}; Do[s0=Prime[n0]; s=0; Do[s+=Prime[n]; If[IntegerQ[s/s0], bb=Append[bb, n-n0]; Break[]], {n, n0+1, 8000}], {n0, 1, 100}]; bb
sncp[n_]:=Module[{p=Prime[n],k=n+1,t},t=Prime[k];While[!Divisible[ t, p], k++;t=t+Prime[k]];k-n]; Array[sncp,100]  (* Harvey P. Dale, May 21 2017 *)

PARI

a(n)=my(p = prime(n), sp = nextprime(p+1), lp = sp, nb = 1); while (sp % p, lp = nextprime(lp+1); nb++; sp += lp); nb; \\ Michel Marcus, May 21 2017

PARI

a(n, p=prime(n))=my(s, k); forprime(q=p+1, , s+=q; k++; if(s%p==0, return(k))) \\ Charles R Greathouse IV, May 21 2017

Extensions

Edited by Don Reble, Nov 10 2005