A112271 One fifth of the sum of the first n primes, when an integer.

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

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, Dec 04 2005 *)
Select[Accumulate[Prime[Range[400]]]/5, IntegerQ] (* Harvey P. Dale, May 03 2017 *)

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