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 A112271 One fifth of the sum of the first n primes, when an integer. 2
 1, 2, 20, 32, 88, 212, 296, 344, 1070, 1166, 1374, 1655, 2248, 2698, 3368, 3730, 3916, 4936, 5160, 5388, 6725, 6983, 8788, 11338, 12382, 12923, 13480, 15026, 16244, 17717, 19033, 19481, 19937, 21108, 24584, 29191, 30345, 33008, 33921, 34850 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 REFERENCES Bach, E. and Shallit, J. Sect. 2.7 in Algorithmic Number Theory, Vol. 1: Efficient Algorithms. Cambridge, MA: MIT Press, 1996. H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206. LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 Leo Moser, Notes on number theory. III. On the sum of consecutive primes, Canad. Math. Bull. 6 (1963), pp. 159-161. Eric Weisstein's World of Mathematics, Prime Sums. FORMULA {a(n)} = {A007504(k)/5 iff 5 | A007504(k)}. {a(n)} = {(p_1 + p_2 + ... + p_k)/5 iff the sum is an integer}. It is sufficient that A007504(k) == 0 (mod 10), but not necessary (the last five consecutive primes ending in 1 can give a solution). It is necessary that k = 2 or k is odd. EXAMPLE a(1) = 1 = (2+3)/5 = A007504(2)/5 = 5/5. a(2) = 2 = (2+3+5)/5 = A007504(3)/5 = 10/5. a(3) = 20 = (2+3+5+7+11+13+17+19+23)/5 = A007504(9)/5 = 100/5. a(4) = 32 = (2+3+5+7+11+13+17+19+23+29+31)/5 = A007504(11)/5 = 160/5. a(5) = 88 = A007504(17)/5 = 440/5. a(6) = 212 = A007504(25)/5 = 1060/5. a(7) = 296 = A007504(29)/5 = 1480/5. a(8) = 344 = A007504(31)/5 = 1720/5. MATHEMATICA s = 0; lst = {}; Do[s = s + Prime[n]; If[Mod[s, 5] == 0, AppendTo[lst, s/5]], {n, 250}]; lst (* Robert G. Wilson v *) Select[Accumulate[Prime[Range[400]]]/5, IntegerQ] (* Harvey P. Dale, May 03 2017 *)112271"] CROSSREFS Cf. A000040, A007504, A111319, A112040. Sequence in context: A268742 A261460 A061471 * A201122 A103076 A293029 Adjacent sequences:  A112268 A112269 A112270 * A112272 A112273 A112274 KEYWORD easy,nonn AUTHOR Jonathan Vos Post, Nov 30 2005 EXTENSIONS More terms from Stefan Steinerberger and Robert G. Wilson v, Dec 04 2005 STATUS approved

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