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%I #21 Mar 16 2023 04:53:08
%S 1,10,2,5,8,49,4,23,23,7,39,29,6,10,39,25,30,151,38,19,139,27,174,21,
%T 287,422,240,24,94,22,16,173,861,231,143,140,213,902,18,134,143,310,
%U 70,58,12,550,237,210,229,57,221,271,194,540,145,718,116,184,90,14,168,455,61,454
%N a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i).
%C 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.
%C 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
%H T. D. Noe, <a href="/A111287/b111287.txt">Table of n, a(n) for n = 1..10000</a>
%H D. K. L. Shiu, <a href="http://dx.doi.org/10.1112/S0024610799007863">Strings of Congruent Primes</a>, J. Lond. Math. Soc. 61 (2) (2000) 359-373 [<a href="http://www.ams.org/mathscinet-getitem?mr=1760689">MR1760689</a>]
%e 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).
%p 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;
%t 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 *)
%o (PARI) A111287(n)= n=Mod(0,prime(n)); for(k=1,1e9, (n+=prime(k)) || return(k)) \\ _M. F. Hasler_, Nov 29 2009
%Y Cf. A000041, A007504, A053050, A111267, A111272, A103208, A168678.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Nov 03 2005
%E The comments are based on correspondence with Paul Pollack and a posting to sci.math by Fred Helenius.
%E Typo in reference fixed by _David Applegate_, Dec 18 2009
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