A249112 -id:A249112 - cp's OEIS frontend

A249113 Take n and successively add 1, 2, ..., a(n) until reaching a prime for the third time.

4, 5, 16, 5, 11, 13, 8, 6, 19, 6, 12, 13, 7, 9, 28, 5, 11, 13, 12, 17, 19, 6, 11, 25, 8, 6, 28, 5, 20, 37, 7, 14, 19, 10, 11, 34, 8, 6, 40, 6, 20, 25, 8, 9, 31, 6, 11, 25, 19, 21, 19, 6, 12, 25, 16, 9, 28, 5, 20, 22, 7, 14, 40, 9, 11, 34, 19, 6, 52, 17, 12

Offset: 1

Gil Broussard, Oct 21 2014

Conjecturally (Hardy & Littlewood conjecture F), a(n) exists for all n. - Charles R Greathouse IV, Oct 21 2014

			a(1)=4 because 1+1+2+3+4=11 and exactly two partial sums are prime (2,7).
a(2)=5 because 2+1+2+3+4+5=17 and exactly two partial sums are prime (3,5).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A085415, A249112.

Table[k = 0; Do[k++; While[! PrimeQ[n + Total@ Range@ k], k++], {x, 3}]; k, {n, 71}] (* Michael De Vlieger, Jan 03 2016 *)

a(n)=my(k,s=3); while(s,if(isprime(n+=k++),s--));k \\ Charles R Greathouse IV, Oct 21 2014

a(n,s=3)=my(k);until(isprime(n+=k++)&&!s--,);k \\ allows one to get A249112(n) as a(n,2). - M. F. Hasler, Oct 21 2014

n+A000217(k) is prime for k=a(n) and exactly two smaller positive values. - M. F. Hasler, Oct 21 2014

A249114 Take the counting numbers and continue adding 1, 2, ..., a(n) until one reaches a fourth prime.

Original entry on oeis.org

7, 6, 19, 10, 12, 25, 11, 9, 40, 13, 15, 25, 11, 17, 67, 6, 15, 22, 15, 18, 43, 9, 12, 34, 12, 9, 31, 9, 32, 58, 8, 21, 28, 14, 12, 37, 11, 9, 55, 13, 23, 46, 11, 14, 43, 10, 15, 34, 24, 26, 28, 9, 15, 37, 23, 18, 40, 6, 24, 61, 8, 18, 43, 22, 27, 37, 20, 9

Offset: 1

Gil Broussard, Oct 21 2014

Conjecturally (Hardy & Littlewood conjecture F), a(n) exists for all n. - Charles R Greathouse IV, Oct 21 2014
It appears that the minimum value reached by a(n) is 6. This occurs for n=2, 16, 58, 136, 178, 418, 598, 808, ... . - Michel Marcus, Oct 26 2014
The conjecture in the previous line is true - if n is odd, then n+1 is even, n+3 is even, n+6 and n+10 are odd, etc., so a(n)>6. If n is even, then +1 and +3 are odd, +6, +10 are even, so the fourth prime can be first for a(n)=6. - Jon Perry, Oct 29 2014
Conjecture: a(n) is odd approximately 50% of the time. - Jon Perry, Oct 29 2014

			a(1) = 7 because 1+1+2+3+4+5+6+7 = 29 and exactly three partial sums are prime (2,7,11).
a(2) = 6 because 2+1+2+3+4+5+6 = 23 and exactly three partial sums are prime (3,5,17).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A085415, A249112, A249113.

f:= proc(n) local j,count;
      count:= 0;
      for j from 1 do
        if isprime(n + j*(j+1)/2) then
           count:= count+1;
           if count = 4 then return j fi
        fi
      od
end proc:
seq(f(n),n=1..100); # Robert Israel, Oct 29 2014

a[n_] := Module[{j, cnt = 0}, For[j = 1, True, j++, If[PrimeQ[n+j(j+1)/2], cnt++; If[cnt == 4, Return[j]]]]];
Array[a, 100] (* Jean-François Alcover, Oct 03 2020, after Maple *)

a(n)=my(k, s=4); while(s, if(isprime(n+=k++), s--)); k \\ Charles R Greathouse IV, Oct 21 2014

a(n) = Min_{k>0 | { n+A000217(j), j=1...k} contains four primes}. - M. F. Hasler, Oct 29 2014

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A249113 Take n and successively add 1, 2, ..., a(n) until reaching a prime for the third time.

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A249114 Take the counting numbers and continue adding 1, 2, ..., a(n) until one reaches a fourth prime.

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Mathematica

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