cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A111287 a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i).

Original entry on oeis.org

1, 10, 2, 5, 8, 49, 4, 23, 23, 7, 39, 29, 6, 10, 39, 25, 30, 151, 38, 19, 139, 27, 174, 21, 287, 422, 240, 24, 94, 22, 16, 173, 861, 231, 143, 140, 213, 902, 18, 134, 143, 310, 70, 58, 12, 550, 237, 210, 229, 57, 221, 271, 194, 540, 145, 718, 116, 184, 90, 14, 168, 455, 61, 454
Offset: 1

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Author

N. J. A. Sloane, Nov 03 2005

Keywords

Comments

It follows from a theorem of Daniel Shiu that k always exists. Shiu has proved that if (a,b) = 1 then the arithmetic progression a, a + b, ..., a + k*b, ... contains arbitrarily long sequences of consecutive primes. Since, for any positive integer b, there are thus arbitrarily long sequences of consecutive primes congruent to 1 mod b, there must be infinitely many a(n) that are divisible by b.
To clarify the previous comment: If the sum of the primes up to some point is s (mod b), then we need exactly b-s consecutive primes equal to 1 (mod b) to produce a sum divisible by b. Hence when there are b-1 consecutive primes congruent to 1 (mod b), then the sum of primes up to one of those primes will be divisible by b. - T. D. Noe, Dec 02 2009

Examples

			A007504 begins 2,5,10,17,28,41,58,77,100,129,... and the k=10th term is the first one that is divisible by prime(2) = 3, so a(2) = 10 (see also A103208).
		

Crossrefs

Programs

  • Maple
    read transforms; M:=1000; p0:=[seq(ithprime(i),i=1..M)]; q0:=PSUM(p0); w:=[]; for n from 1 to M do p:=p0[n]; hit := 0; for i from 1 to M do if q0[i] mod p = 0 then w:=[op(w),i]; hit:=1; break; fi; od: if hit = 0 then break; fi; od: w;
  • Mathematica
    Table[p=Prime[n]; s=0; k=0; While[k++; s=Mod[s+Prime[k],p]; s>0]; k, {n,10}] (* T. D. Noe, Dec 02 2009 *)
  • PARI
    A111287(n)= n=Mod(0,prime(n)); for(k=1,1e9, (n+=prime(k)) || return(k)) \\ M. F. Hasler, Nov 29 2009

Extensions

The comments are based on correspondence with Paul Pollack and a posting to sci.math by Fred Helenius.
Typo in reference fixed by David Applegate, Dec 18 2009