A337802 Minimum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.

0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Andres Cicuttin, Sep 22 2020

Keywords

• nonn

Comments

In the first 1000 terms, a(n) = 1 only for n = 3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 47, 61, 73, 107, 109, 113, 181, 199, and 467.

Is there an index k such that a(n) = 0 for n > k ?

Examples

			The primes among the first 5 positive integers (1,2,3,4,5) are 2, 3, and 5, then the corresponding characteristic function of primes is (0,1,1,0,1) (see A010051) and the corresponding five possible cyclic self-convolutions are the dot products between (0,1,1,0,1) and the rotations of its mirrored version as shown below:
  (0,1,1,0,1).(1,0,1,1,0) = 0*1 + 1*0 + 1*1 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,0,1,1) = 0*0 + 1*1 + 1*0 + 0*1 + 1*1 = 2,
  (0,1,1,0,1).(1,0,1,0,1) = 0*1 + 1*0 + 1*1 + 0*0 + 1*1 = 2,
  (0,1,1,0,1).(1,1,0,1,0) = 0*1 + 1*1 + 1*0 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,1,0,1) = 0*0 + 1*1 + 1*1 + 0*0 + 1*1 = 3.
Then a(5)=1 because 1 is the minimum among the five values.
		

Crossrefs

Cf. A337327, A010051, A299111, A014342.

Programs

Mathematica

b[n_] := Table[If[PrimeQ[i], 1, 0], {i, 1, n}];
Table[Min@Table[b[n].RotateRight[Reverse[b[n]], j], {j, 0, n - 1}], {n, 1, 100}]

PARI

a(n) = vecmin(vector(n, k, sum(i=1, n, isprime(n-i+1)*isprime(1+(i+k)%n)))); \\ Michel Marcus, Sep 23 2020