A337327 - OEIS

A337327

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

0, 1, 2, 2, 3, 2, 4, 3, 3, 3, 4, 4, 6, 5, 4, 4, 6, 5, 8, 7, 6, 6, 8, 7, 8, 7, 6, 6, 8, 7, 10, 9, 8, 8, 8, 8, 10, 9, 8, 8, 10, 10, 12, 12, 10, 11, 12, 12, 12, 13, 12, 12, 14, 13, 14, 13, 12, 12, 12, 12, 14, 13, 12, 13, 12, 14, 14, 15, 12, 14, 14, 16, 16, 18

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OFFSET

1,3

LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Andres Cicuttin, Graph of first 2^10 terms

EXAMPLE

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)=3 because 3 is the maximum among the five values.

MATHEMATICA

b[n_]:=Table[If[PrimeQ[i], 1, 0], {i, 1, n}];

Table[Max@Table[b[n].RotateRight[Reverse[b[n]], j], {j, 0, n-1}], {n, 1, 100}]

PROG

(PARI) a(n) = vecmax(vector(n, k, sum(i=1, n, isprime(n-i+1)*isprime(1+(i+k)%n)))); \\ Michel Marcus, Aug 26 2020

CROSSREFS