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A051709
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a(n) = sigma(n) + phi(n) - 2n.
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17
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0, 0, 0, 1, 0, 2, 0, 3, 1, 2, 0, 8, 0, 2, 2, 7, 0, 9, 0, 10, 2, 2, 0, 20, 1, 2, 4, 12, 0, 20, 0, 15, 2, 2, 2, 31, 0, 2, 2, 26, 0, 24, 0, 16, 12, 2, 0, 44, 1, 13, 2, 18, 0, 30, 2, 32, 2, 2, 0, 64, 0, 2, 14, 31, 2, 32, 0, 22, 2, 28, 0, 75, 0, 2, 14, 24, 2, 36, 0
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OFFSET
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1,6
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COMMENTS
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Because sigma and phi are multiplicative functions, it is easy to show that (1) if a(n)=0, then n is prime or 1 and (2) if a(n)=2, then n is the product of two distinct prime numbers. Note that a(n) is the n-th term of the Dirichlet series whose generating function is given below. Using the generating function, it is theoretically possible to compute a(n). Hence a(n)=0 could be used as a primality test and a(n)=2 could be used as a test for membership in P2 (A006881). - T. D. Noe, Aug 01 2002
It appears that a(n) - A002033(n) = zeta(s-1) * (zeta(s) - 2 + 1/zeta(s)) + 1/(zeta(s)-2). - Eric Desbiaux, Jul 04 2013
a(n) = 1 if and only if n = prime(k)^2 (n is in A001248). It seems that a(n) = k has only finitely many solutions for k >= 3. - Jianing Song, Jun 27 2021
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LINKS
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FORMULA
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Dirichlet g.f.: zeta(s-1) * (zeta(s) - 2 + 1/zeta(s)). - T. D. Noe, Aug 01 2002
a(n) = A001065(n) - A051953(n). [Difference between the sum of proper divisors of n and their Moebius-transform.]
Sum_{k=1..n} a(k) = (3/(Pi^2) + Pi^2/12 - 1) * n^2 + O(n*log(n)). - Amiram Eldar, Dec 03 2023
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EXAMPLE
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a(5) = sigma(5) + phi(5) - 2*5 = 6 + 4 - 10 = 0.
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MATHEMATICA
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Table[DivisorSigma[1, n]+EulerPhi[n]-2n, {n, 80}] (* Harvey P. Dale, Apr 08 2015 *)
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PROG
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CROSSREFS
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Cf. A000010, A000203, A001065, A001248, A005843, A006881, A051612, A051953, A065387, A072780, A228498 (= a(n^2)), A297159, A324048, A344994, A344995, A344996, A345048, A345054.
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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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