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%I #15 Feb 17 2022 10:01:42
%S 1,2,20,32,88,212,296,344,1070,1166,1374,1655,2248,2698,3368,3730,
%T 3916,4936,5160,5388,6725,6983,8788,11338,12382,12923,13480,15026,
%U 16244,17717,19033,19481,19937,21108,24584,29191,30345,33008,33921,34850
%N One fifth 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="/A112271/b112271.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)/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.
%e a(1) = 1 = (2+3)/5 = A007504(2)/5 = 5/5.
%e a(2) = 2 = (2+3+5)/5 = A007504(3)/5 = 10/5.
%e a(3) = 20 = (2+3+5+7+11+13+17+19+23)/5 = A007504(9)/5 = 100/5.
%e a(4) = 32 = (2+3+5+7+11+13+17+19+23+29+31)/5 = A007504(11)/5 = 160/5.
%e a(5) = 88 = A007504(17)/5 = 440/5.
%e a(6) = 212 = A007504(25)/5 = 1060/5.
%e a(7) = 296 = A007504(29)/5 = 1480/5.
%e a(8) = 344 = A007504(31)/5 = 1720/5.
%t s = 0; lst = {}; Do[s = s + Prime[n]; If[Mod[s, 5] == 0, AppendTo[lst, s/5]], {n, 250}]; lst (* _Robert G. Wilson v_ *)
%t Select[Accumulate[Prime[Range[400]]]/5,IntegerQ] (* _Harvey P. Dale_, May 03 2017 *)112271"]
%Y Cf. A000040, A007504, A111319, A112040.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Nov 30 2005
%E More terms from _Stefan Steinerberger_ and _Robert G. Wilson v_, Dec 04 2005
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