

A337802


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


2



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
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OFFSET

1,1


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 ?


LINKS

Table of n, a(n) for n=1..106.


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)=1 because 1 is the minimum among the five values.


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}]


PROG

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


CROSSREFS

Cf. A337327, A010051, A299111, A014342.
Sequence in context: A219071 A072629 A164292 * A257531 A151763 A324908
Adjacent sequences: A337799 A337800 A337801 * A337803 A337804 A337805


KEYWORD

nonn


AUTHOR

Andres Cicuttin, Sep 22 2020


STATUS

approved



