A305875 a(n) is the maximum number m such that 2n = prime(i+k) + prime(j-k), k=1,2,...,m, where i and j are a pair of numbers such that 2n = prime(i+1) + prime(j+1).

1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 3, 1, 1, 3, 1, 2, 2, 1, 1, 2, 1, 2, 3, 1, 1, 4, 1, 1, 5, 1, 2, 2, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 6, 1, 1, 2, 2, 1, 4, 1, 2, 3, 2, 2, 5, 1, 1, 6, 1, 1, 4, 1, 1, 3, 1, 2, 2, 2, 2, 3, 1, 1, 4, 1, 2, 2, 1, 2, 2, 1, 1, 4, 1, 1, 2, 1

Offset

2,4

Comments

Conjecture: there are only 392 n's such that a(n) = 1.

Links

Table of n, a(n) for n=2..88.

Example

For n=2, 2n=4 = 2+2, this is the only case, so a(2)=1;
For n=3, 2n=6 = 3+3, this is the only case, so a(3)=1;
...
For n=12, 2n=24 = 5+19 = 7+17 = 11+13, both {3,5,7} and {19,17,13} are consecutive prime lists with length 3, so a(12)=3;
...
For n=33, 2n=66 = 5+61 = 7+59 = 13+53 = 19+47 = 23+43 = 29+37, both {5,7} and {61, 59} are consecutive prime lists with length 2, and although {19,23,29} is a consecutive prime list with length 3, its counterpart {47,43,37} is not a consecutive prime list, so a(33)=2 but not 3.

Mathematica

Table[maxct = 0; ct = 0; strike = 0; p = 1; pbuf = 1;
While[p = NextPrime[p]; (2*p) <= i,
  If[PrimeQ[i - p],
   If[strike == 0, ct = 1; pbuf = i - p; strike = 1,
    If[pbuf == NextPrime[i - p], ct++; pbuf = i - p, strike = 0;
     If[maxct < ct, maxct = ct]]], strike = 0;
   If[maxct < ct, maxct = ct]]];
If[maxct < ct, maxct = ct]; maxct, {i, 4, 176, 2}]

Crossrefs

Cf. A002375, A045917, A002372.

Sequence in context: A185090 A115268 A103610 * A109705 A352578 A278341

Adjacent sequences: A305872 A305873 A305874 * A305876 A305877 A305878

Keyword

nonn,easy

Author

Lei Zhou, Jun 12 2018

Status

approved