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Search: id:A111287
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| A111287 |
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a(n) = smallest k such that prime(n) divides Sum_{i=1..k} prime(i). |
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+0
7
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| 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
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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.
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REFERENCES
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D. K. L. Shiu, Strings of congruent primes, J. London Math. Soc. 61 (2000), 359-373; MR 2001f:11155.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000
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EXAMPLE
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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).
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MAPLE
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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;
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CROSSREFS
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Cf. A000041, A007504, A053050, A111267, A111272, A103208, etc.
Sequence in context: A069036 A155817 A037922 this_sequence A084455 A069532 A084461
Adjacent sequences: A111284 A111285 A111286 this_sequence A111288 A111289 A111290
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Nov 03 2005
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EXTENSIONS
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The comments are based on correspondence with Paul Pollack and a posting to sci.math by Fred Helenius.
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