OFFSET
1,1
COMMENTS
Conjecture: a(n) exists for every n.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
a(3)=6 because prime(3)=5 divides 7+11+13+17+19+23 = 90.
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Zak Seidov, Jul 20 2003
EXTENSIONS
Edited by Don Reble, Nov 10 2005
STATUS
approved