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%I #12 Aug 24 2022 08:51:21
%S 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,
%T 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,
%U 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
%N Minimum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.
%C 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.
%C Is there an index k such that a(n) = 0 for n > k ?
%e 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:
%e (0,1,1,0,1).(1,0,1,1,0) = 0*1 + 1*0 + 1*1 + 0*1 + 1*0 = 1,
%e (0,1,1,0,1).(0,1,0,1,1) = 0*0 + 1*1 + 1*0 + 0*1 + 1*1 = 2,
%e (0,1,1,0,1).(1,0,1,0,1) = 0*1 + 1*0 + 1*1 + 0*0 + 1*1 = 2,
%e (0,1,1,0,1).(1,1,0,1,0) = 0*1 + 1*1 + 1*0 + 0*1 + 1*0 = 1,
%e (0,1,1,0,1).(0,1,1,0,1) = 0*0 + 1*1 + 1*1 + 0*0 + 1*1 = 3.
%e Then a(5)=1 because 1 is the minimum among the five values.
%t b[n_] := Table[If[PrimeQ[i], 1, 0], {i, 1, n}];
%t Table[Min@Table[b[n].RotateRight[Reverse[b[n]], j], {j, 0, n - 1}], {n, 1, 100}]
%o (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
%Y Cf. A337327, A010051, A299111, A014342.
%K nonn
%O 1,1
%A _Andres Cicuttin_, Sep 22 2020
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