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A089496
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a(n) = mu(prime(n)+1) + mu(prime(n)-1), where mu is the Moebius function.
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4
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0, -1, 1, 1, 1, 1, 0, 0, 1, -1, -1, 1, -1, -1, 1, 0, 1, 1, -1, -1, 1, -1, 1, 0, 0, -1, -1, 1, -1, -1, 0, -1, -1, -1, 0, 0, 1, 0, 1, -1, 1, -1, -1, 1, 0, 0, 1, -1, 1, -1, 0, -1, 0, 0, -1, 1, 0, 0, 1, -1, -1, 0, 0, -1, 1, -1, 1, 0, 1, 0, -1, 1, -1, -1, 0, 1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 0, 1, 1, 1, 1, 1, 0, 0, -1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, -1
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OFFSET
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1
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COMMENTS
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This sum is always -1, 0 or 1 because for odd prime p, both p-1 and p+1 cannot be squarefree; one of them will be divisible by 4. This also implies that terms in this sequence are zero only for 2 and odd primes p such that mu(p-1) = mu(p+1) = 0, which is A075432.
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LINKS
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FORMULA
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Let p = prime(n), then a(n) = mu(p+(-1/p)), where (-1/p) is the Legendre symbol, A070750. (Pieter Moree). (This is true for n > 1) - Antti Karttunen, Jul 23 2017
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MATHEMATICA
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Table[MoebiusMu[Prime[n]+1] + MoebiusMu[Prime[n]-1], {n, 1, 150}]
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PROG
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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