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A320626
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a(n) = (A006134(prime(n)-1) - Legendre(prime(n), 3))/prime(n)^2.
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1
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1, 1, 4, 26, 2074, 21660, 2804068, 33847970, 5345496688, 12201269878660, 165029257057602, 433037204976615540, 85637003420093445994, 1215603499085916654728, 248871244937134915010626, 753881874165723132009014662, 2359161060012378685209851991166
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OFFSET
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1,3
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COMMENTS
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a(n) is always an integer.
Are there any primes p such that p^3 divides A006134(p-1) - Legendre(p, 3)?
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LINKS
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EXAMPLE
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a(1) = (binomial(0, 0) + binomial(2, 1) + 1)/4 = 4/4 = 1.
a(2) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2))/9 = 9/9 = 1.
a(3) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + binomial(6, 3) + binomial(8, 4) + 1)/25 = 100/25 = 4.
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PROG
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(PARI) A006134(n) = sum(k=0, n, binomial(2*k, k))
a(n) = my(p=prime(n)); (A006134(p-1) - kronecker(p, 3))/p^2
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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