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%I #35 Aug 24 2022 08:51:26
%S 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,
%T 8,8,10,9,8,8,10,10,12,12,10,11,12,12,12,13,12,12,14,13,14,13,12,12,
%U 12,12,14,13,12,13,12,14,14,15,12,14,14,16,16,18
%N Maximum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.
%H Rémy Sigrist, <a href="/A337327/b337327.txt">Table of n, a(n) for n = 1..10000</a>
%H Andres Cicuttin, <a href="/A337327/a337327.pdf">Graph of first 2^10 terms</a>
%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)=3 because 3 is the maximum among the five values.
%t b[n_]:=Table[If[PrimeQ[i],1,0],{i,1,n}];
%t Table[Max@Table[b[n].RotateRight[Reverse[b[n]],j],{j,0,n-1}],{n,1,100}]
%o (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
%Y Cf. A010051, A299111, A014342.
%K nonn,look
%O 1,3
%A _Andres Cicuttin_, Aug 23 2020
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