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A337802
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Minimum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.
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2
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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
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1,1
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COMMENTS
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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 ?
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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