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%I #27 Feb 07 2018 16:31:24
%S 4,12,31,62,133,224,377,558,865,1304,1805,2462,3337,4280,5389,6726,
%T 8449,10264,12663,15294,18061,21200,24961,29166,34173,39508,45017,
%U 50870,57141,63788,72299,81234,91365,101732,113327,125166,138355,152348,167179,182862
%N Minimum value of the cyclic autocorrelation of first n primes.
%C Maximum values of the cyclic autocorrelation of first n primes are in A024450.
%C If we use this definition with integers instead of primes it is obtained A088003.
%H Alois P. Heinz, <a href="/A299053/b299053.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = Min_{k=1..n} Sum_{i=1..n} prime(i)*prime(1 + (i+k) mod n).
%e For n = 4 the four possible cyclic autocorrelations of first four primes are:
%e (2,3,5,7).(2,3,5,7) = 2*2 + 3*3 + 5*5 + 7*7 = 4 + 9 + 25 + 49 = 87,
%e (2,3,5,7).(7,2,3,5) = 2*7 + 3*2 + 5*3 + 7*5 = 14 + 6 + 15 + 35 = 70,
%e (2,3,5,7).(5,7,2,3) = 2*5 + 3*7 + 5*2 + 7*3 = 10 + 21 + 10 + 21 = 62,
%e (2,3,5,7).(3,5,7,2) = 2*3 + 3*5 + 5*7 + 7*2 = 6 + 15 + 35 + 14 = 70,
%e then a(4)=62 because 62 is the minimum among the four values.
%p a:= n-> min(seq(add(ithprime(i)*ithprime(irem(i+k, n)+1), i=1..n), k=1..n)):
%p seq(a(n), n=1..40); # _Alois P. Heinz_, Feb 06 2018
%t p[n_]:=Prime[Range[n]];
%t Table[Table[p[n].RotateRight[p[n],j],{j,0,n-1}]//Min,{n,1,36}]
%o (PARI) a(n) = vecmin(vector(n, k, sum(i=1, n, prime(i)*prime(1+(i+k)%n)))); \\ _Michel Marcus_, Feb 07 2018
%Y Cf. A024450, A014342, A299111, A088003.
%K nonn
%O 1,1
%A _Andres Cicuttin_, Feb 01 2018
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