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Search: id:A002173
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| A002173 |
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Sum_{d|n, d == 1 mod 4} d^2 - Sum_{d|n, d == 3 mod 4} d^2.
(Formerly M4467 N1895)
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+0
5
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| 1, 1, -8, 1, 26, -8, -48, 1, 73, 26, -120, -8, 170, -48, -208, 1, 290, 73, -360, 26, 384, -120, -528, -8, 651, 170, -656, -48, 842, -208, -960, 1, 960, 290, -1248, 73, 1370, -360, -1360, 26, 1682, 384, -1848, -120, 1898, -528, -2208, -8, 2353, 651, -2320, 170
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Multiplicative because it is the Inverse Moebius transform of [1 0 -3^2 0 5^2 0 -7^2 ...], which is multiplicative. Christian G. Bower (bowerc(AT)usa.net) May 18, 2005.
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
J. W. L. Glaisher, On the function chi(n), Quarterly Journal of Pure and Applied Mathematics, 20 (1884), 97-167.
N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 85, Eq. (32.7).
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LINKS
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J. Stienstra, Mahler measure, Eisenstein series and dimers
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FORMULA
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Multiplicative with a(p^e) = 1 if p = 2; ((p^2)^(e+1)-1)/(p^2-1) if p == 1 (mod 4); ((-p^2)^(e+1)-1)/(-p^2-1) if p == 3 (mod 4). - David W. Wilson, Sep 01, 2001
G.f.: Sum[n>=1, A056594(n-1)*n^2*q^n/(1-q^n) ].
Expansion of (1-theta_4(q)^2*theta_4(q^2)^4)/4 in powers of q. - Michael Somos Aug 09 2006
Expansion of (1-eta(q)^4*eta(q^2)^6/eta(q^4)^4)/4 in powers of q.
G.f.: qG'(q)/G(q), with G(q) = Prod[n>=1, (1-q^n)^(4n*A056594(n+1)) ].
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EXAMPLE
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The divisors of 15 are 1,3,5,15, so a(15)=(1^2+5^2)-(3^2+15^2) = -208.
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PROGRAM
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(PARI) {a(n)=if(n<1, 0, sumdiv(n, d, d^2*kronecker(-4, d)))} /* Michael Somos Aug 09 2006 */
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CROSSREFS
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Equals A050450(n) - A050453(n).
A120030(n)=-4*a(n), if n>0.
Sequence in context: A125235 A019432 A138505 this_sequence A050458 A125166 A075155
Adjacent sequences: A002170 A002171 A002172 this_sequence A002174 A002175 A002176
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KEYWORD
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sign,mult
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
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More terms from David W. Wilson (davidwwilson(AT)comcast.net).
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