

A337327


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


2



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 selfconvolutions are the dot products between (0,1,1,0,1) and the rotations of its mirrored version as shown here 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, n1}], {n, 1, 100}]


PROG

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


CROSSREFS

Cf. A010051, A299111, A014342.
Sequence in context: A163870 A327664 A155043 * A065770 A297113 A086375
Adjacent sequences: A337324 A337325 A337326 * A337328 A337329 A337330


KEYWORD

nonn,look


AUTHOR

Andres Cicuttin, Aug 23 2020


STATUS

approved



