%I #32 May 14 2024 03:59:11

%S 10,16,18,20,24,26,28,30,32,34,36,40,42,44,46,52,54,57,68,70,74,76,78,

%T 80,82,84,86,88,90,97,99,103,105,107,111,113,119,121,123,125,127,129,

%U 134,136,138,161,163,166,169,175,177,179,185,187,195,197,199,203,205,207,211,213

%N Numbers k such that 3 divides prime(1) + ... + prime(k).

%C Also, numbers k such that 3 divides the concatenation of the first k primes (see A019518).

%C The first comment and the description are true whenever the number of primes congruent to 1 mod 6 exceeds the number of primes congruent to 5 mod 6 and the difference is congruent to 1 mod 3 or the number of primes congruent to 5 mod 6 exceeds the number of primes congruent to 1 mod 6 and the difference is congruent to 2 mod 3. - _Roderick MacPhee_, Oct 30 2015

%H Amiram Eldar, <a href="/A103208/b103208.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Harvey P. Dale)

%H Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/sm_prime.htm">Smarandache consecutive prime sequences (n = 1 to 100)</a>.

%p s1:=[2]; M:=1000; for n from 2 to M do s1:=[op(s1),s1[n-1]+ithprime(n)]; od: s1;

%p f:=proc(k) global M,s1; local t1,n; t1:=[]; for n from 1 to M do if s1[n] mod k = 0 then t1:=[op(t1),n]; fi; od: t1; end; f(3);

%t f[n_] := FromDigits[ Flatten[ Table[ IntegerDigits[ Prime[i]], {i, n}]]]; Select[ Range[ 206], Mod[f[ # ], 3] == 0 &]

%t Flatten[Position[Accumulate[Prime[Range[250]]],_?(Divisible[#,3]&)]] (* _Harvey P. Dale_, Jan 14 2016 *)

%o (PARI) a=0;b=0;for(x=3,1000,if(prime(x)%6==1,a+=1,b+=1);if((a-b)%3==1 || (b-a)%3==2,print1(x","))) \\ _Roderick MacPhee_, Oct 30 2015

%o (PARI) lista(nn) = { s=0; for(k=1, nn, s += prime(k); if(s % 3 == 0, print1(k, ", ")););} \\ _Altug Alkan_, Dec 04 2015

%Y Cf. A007504, A019518, A104644, A111287.

%Y Cf. A111318, A111319, A111320, A111321, A111322, A111323, A111324, A111325, A111326, A111327.

%K nonn

%O 1,1

%A _Robert G. Wilson v_, Mar 19 2005

%E Entry revised by _N. J. A. Sloane_, Nov 09 2005