A381869 Smallest starting prime for which the sum of 2*n consecutive primes is 0 modulo 10, or -1 if no such prime exists.

13, 11, 7, 7, 13, 17, 7, 17, 37, 3, 7, 41, 7, 7, 11, 11, 11, 11, 11, 13, 11, 13, 11, 7, 7, 17, 7, 43, 41, 3, 3, 13, 11, 7, 13, 19, 7, 11, 11, 29, 7, 43, 3, 7, 11, 13, 23, 29, 3, 7, 7, 11, 11, 11, 19, 13, 5, 5, 13, 37, 17, 3, 3, 7, 17, 17, 3, 11, 19, 13, 3, 7, 23

Offset: 1

Author

Jean-Marc Rebert, Mar 09 2025

Keywords

• nonn

Examples

			a(1) = 13, because 13 and 17 are 2*1 = 2 consecutive primes such that 13 + 17 = 20 and 20 modulo 10 = 0, and no smaller prime has this property.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007652, A111324.

Programs

Maple

P:= select(isprime,[2,seq(i,i=3..10^6,2)]):
S:= ListTools:-PartialSums(P):
f:= proc(n) local j,t;
  for j from 1 do
    if S[2*n+j] - S[j] mod 10 = 0 then return P[j+1] fi
  od
end proc:
map(f, [$1..100]); # Robert Israel, May 08 2025

Mathematica

Do[i=1;Until[Mod[Total[Prime[Range[i,i+2*n-1]]],10]==0,i++];a[n]=Prime[i],{n,73}];Array[a,73] (* James C. McMahon, Mar 23 2025 *)

PARI

isok(p, n) = my(i=primepi(p), q=prime(2*n+i-1)); vecsum(apply(x->Mod(x,10), primes([p, q]))) == 0;
a(n) = my(p=3); while (!isok(p, n), p=nextprime(p+1)); p; \\ Michel Marcus, Mar 09 2025

Python

from sympy import nextprime, prime, sieve
def a(n):
    plst = list(sieve.primerange(3, prime(2*n+1)+1))
    s = sum(plst)
    while s%10:
        q = nextprime(plst[-1])
        s += (q-plst[0])
        plst = plst[1:] + [q]
    return plst[0]
print([a(n) for n in range(1, 74)]) # Michael S. Branicky, Mar 09 2025