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a(n) = (A006134((prime(n)-1)/2) - Legendre(prime(n), 3))/prime(n).
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%I #8 Oct 26 2018 00:56:39

%S 1,2,4,32,98,1034,3484,41582,1868198,6751460,330246410,4522217312,

%T 16829506610,235344648416,12559206444236,683534482499900,

%U 2599887148071402,144419015585768252,2117118216002111198,8120930597438173620

%N a(n) = (A006134((prime(n)-1)/2) - Legendre(prime(n), 3))/prime(n).

%C a(n) is always an integer.

%C Primes p such that p^2 divides A006134((p-1)/2) - Legendre(p, 3) are p = 103, ... What's the next?

%F a(2) = (binomial(0, 0) + binomial(2, 1))/3 = 3/3 = 1.

%F a(3) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + 1)/5 = 10/5 = 2.

%F a(4) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + binomial(6, 3) - 1)/7 = 28/7 = 4.

%o (PARI) A006134(n) = sum(k=0, n, binomial(2*k,k))

%o a(n) = my(p=prime(n)); (A006134((p-1)/2) - kronecker(p,3))/p

%Y Cf. A006134, A102283, A320626.

%K nonn

%O 2,2

%A _Jianing Song_, Oct 18 2018