A337327 Maximum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.
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
Offset: 1
Examples
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)=3 because 3 is the maximum among the five values.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Andres Cicuttin, Graph of first 2^10 terms
Programs
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Mathematica
b[n_]:=Table[If[PrimeQ[i],1,0],{i,1,n}]; Table[Max@Table[b[n].RotateRight[Reverse[b[n]],j],{j,0,n-1}],{n,1,100}]
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PARI
a(n) = vecmax(vector(n, k, sum(i=1, n, isprime(n-i+1)*isprime(1+(i+k)%n)))); \\ Michel Marcus, Aug 26 2020