Search: id:A111287 Results 1-1 of 1 results found. %I A111287 %S A111287 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, %T A111287 422,240,24,94,22,16,173,861,231,143,140,213,902,18,134,143,310,70,58, 12, %U A111287 550,237,210,229,57,221,271,194,540,145,718,116,184,90,14,168,455,61,454 %N A111287 a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i). %C A111287 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. %D A111287 D. K. L. Shiu, Strings of congruent primes, J. London Math. Soc. 61 (2000), 359-373; MR 2001f:11155. %H A111287 T. D. Noe, Table of n, a(n) for n=1..1000 %e A111287 A007504 begins 2,5,10,17,28,41,58,77,100,129,... and the k=10-th term is the first one that is divisible by prime(2) = 3, so a(2) = 10 (see also A103208). %p A111287 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; %Y A111287 Cf. A000041, A007504, A053050, A111267, A111272, A103208, etc. %Y A111287 Sequence in context: A069036 A155817 A037922 this_sequence A084455 A069532 A084461 %Y A111287 Adjacent sequences: A111284 A111285 A111286 this_sequence A111288 A111289 A111290 %K A111287 nonn %O A111287 1,2 %A A111287 N. J. A. Sloane (njas(AT)research.att.com), Nov 03 2005 %E A111287 The comments are based on correspondence with Paul Pollack and a posting to sci.math by Fred Helenius. Search completed in 0.001 seconds