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 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 (list; graph; refs; listen; history; text; internal format)
 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 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 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(n-i+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

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Last modified July 29 03:57 EDT 2021. Contains 346340 sequences. (Running on oeis4.)