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%I #14 Feb 17 2022 10:01:36
%S 43,127,167,213,321,387,457,531,617,709,809,1029,1149,1277,1409,1863,
%T 2027,2290,3397,3629,4113,4367,4629,4899,5179,5467,5761,6063,6371,
%U 7516,7864,8600,8980,9368,10168,10578,11856,12296,12746,13204,13674,14156
%N One third of the sum of the first n primes, when an integer.
%D Bach, E. and Shallit, J. Sect. 2.7 in Algorithmic Number Theory, Vol. 1: Efficient Algorithms. Cambridge, MA: MIT Press, 1996.
%D H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.
%H Harvey P. Dale, <a href="/A112270/b112270.txt">Table of n, a(n) for n = 1..1000</a>
%H Leo Moser, <a href="https://doi.org/10.4153/CMB-1963-013-1">Notes on number theory. III. On the sum of consecutive primes</a>, Canad. Math. Bull. 6 (1963), pp. 159-161.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeSums.html">Prime Sums</a>.
%F {a(n)} = {A007504(k)/3 iff 3 | A007504(k)}. {a(n)} = {(p_1 + p_2 + ... + p_k)/3 iff the sum is an integer}. It is necessary but not sufficient for k to be even.
%e a(1) = 43 = (2+3+5+7+11+13+17+19+23+29)/3 = A007504(10)/3 = 129/3.
%e a(2) = 127 = A007504(16)/3 = 381/3.
%e a(3) = 167 = A007504(18)/3 = 501/3.
%e a(4) = 213 = A007504(20)/3 = 639/3.
%e a(5) = 321 = A007504(24)/3 = 963/3.
%e a(6) = 387 = A007504(26)/3 = 1161/3.
%t s = 0; lst = {}; Do[s = s + Prime[n]; If[Mod[s, 3] == 0, AppendTo[lst, s/3]], {n, 130}]; lst (* _Robert G. Wilson v_ *)
%t Select[Accumulate[Prime[Range[200]]]/3,IntegerQ] (* _Harvey P. Dale_, Feb 20 2018 *)
%Y Cf. A000040, A007504, A112040.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Nov 30 2005
%E More terms from _Robert G. Wilson v_, Nov 30 2005
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