A318351 a(n) is the smallest prime p such that the sum of the first 2*n + 1 odd primes starting with p is prime.

3, 5, 5, 17, 3, 5, 29, 3, 3, 11, 7, 7, 5, 7, 13, 13, 7, 5, 5, 13, 7, 7, 7, 7, 11, 17, 3, 3, 97, 29, 3, 13, 3, 19, 19, 3, 5, 3, 23, 7, 11, 53, 31, 89, 53, 19, 11, 3, 17, 23, 83, 11, 5, 47, 37, 5, 17, 3, 3, 29, 23, 5, 5, 5, 59, 7, 7, 31, 3, 67, 3, 3, 89, 71, 31, 41, 29

Offset: 0

Author

David James Sycamore, Aug 24 2018

Keywords

• nonn

Comments

Conjecture: Sequence is bounded.

The sum of consecutive odd primes is the difference of two terms of A007504, which might be used to find terms for this sequence. - David A. Corneth, Aug 25 2018

Apart from the first term the same as A089793. - R. J. Mathar, Nov 02 2018

Examples

			a(1) = 5 because 3 + 5 + 7 = 15 but 5 + 7 + 11 = 23.
From _David A. Corneth_, Sep 04 2018: (Start)
Partial sums of the primes is sequence A007504; 2, 5, 10, 17, 28, 41, 58, 77, 100, 129, 160, 197, ...
For n = 1, the least k such that A007504(k + 2*n + 1) - A007504(k) is prime is at k = 2 so a(1) is prime(k + 1) = prime(3) = 5.
(End)
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000040, A007504, A065091, A071148.

Programs

Maple

N:= 100: # to get a(0)..a(N)
Primes:= [0,seq(ithprime(i),i=2..5/2*N)]:
PS:= ListTools:-PartialSums(Primes):
found:= true:
for n from 0 to 100 while found do
  found:= false;
  for k from 1 to 5/2*N - (2*n+1) do
    if isprime(PS[k+2*n+1]-PS[k]) then
      found:= true; A[n]:= Primes[k+1]; break
    fi
  od
od:
seq(A[n],n=0..N); # Robert Israel, Oct 21 2018

Mathematica

Array[Block[{k = 1}, While[! PrimeQ@ Total@ Prime[k + Range[2 # + 1]], k++]; Prime[k + 1]] &, 77, 0] (* Michael De Vlieger, Aug 25 2018 *)

PARI

a(n) = {c = 2*n + 1; t=2; while(!isprime(sum(i = t, t + c - 1, prime(i))), t++); prime(t)} \\ David A. Corneth, Sep 04 2018

PARI

upto(n) = {c = n<<1; c += (1-c%2); my(primeSums = List([3]), res = List([3])); t=0; forprime(p = 3, prime(c), t++; listput(primeSums, primeSums[t] + p)); forstep(i = 3, #primeSums, 2, for(j = 1, #primeSums - i,   if(isprime(primeSums[i + j] - primeSums[j]), listput(res,  primeSums[j+1] - primeSums[j]); next(2)))); res} \\ gives at most the first n terms \\ David A. Corneth, Sep 04 2018