A103208 Numbers k such that 3 divides prime(1) + ... + prime(k).

10, 16, 18, 20, 24, 26, 28, 30, 32, 34, 36, 40, 42, 44, 46, 52, 54, 57, 68, 70, 74, 76, 78, 80, 82, 84, 86, 88, 90, 97, 99, 103, 105, 107, 111, 113, 119, 121, 123, 125, 127, 129, 134, 136, 138, 161, 163, 166, 169, 175, 177, 179, 185, 187, 195, 197, 199, 203, 205, 207, 211, 213

Offset: 1

Author

Robert G. Wilson v, Mar 19 2005

Keywords

• nonn

Comments

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

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

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
• Hisanori Mishima, Smarandache consecutive prime sequences (n = 1 to 100).

Crossrefs

Cf. A007504, A019518, A104644, A111287.

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

Programs

Maple

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

Mathematica

f[n_] := FromDigits[ Flatten[ Table[ IntegerDigits[ Prime[i]], {i, n}]]]; Select[ Range[ 206], Mod[f[ # ], 3] == 0 &]
Flatten[Position[Accumulate[Prime[Range[250]]],?(Divisible[#,3]&)]] (* _Harvey P. Dale, Jan 14 2016 *)

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

PARI

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

Extensions

Entry revised by N. J. A. Sloane, Nov 09 2005