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a(n) = tau(n)^2, where tau(n) = A000005(n).
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%I #49 Sep 08 2022 08:44:52

%S 1,4,4,9,4,16,4,16,9,16,4,36,4,16,16,25,4,36,4,36,16,16,4,64,9,16,16,

%T 36,4,64,4,36,16,16,16,81,4,16,16,64,4,64,4,36,36,16,4,100,9,36,16,36,

%U 4,64,16,64,16,16,4,144,4,16

%N a(n) = tau(n)^2, where tau(n) = A000005(n).

%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, Theorem 304.

%H T. D. Noe, <a href="/A035116/b035116.txt">Table of n, a(n) for n = 1..1000</a>

%H Mircea Merca, <a href="https://www.researchgate.net/publication/312324402">The Lambert series factorization theorem</a>, The Ramanujan Journal, January 2017; DOI: 10.1007/s11139-016-9856-3.

%F Dirichlet g.f.: zeta(s)^4/zeta(2s).

%F tau(n)^2 = Sum_{d|n} tau(d^2), Dirichlet convolution of A048691 and A000012 (i.e.: inverse Mobius transform of A048691).

%F Multiplicative with a(p^e) = (e+1)^2. - _Vladeta Jovovic_, Dec 03 2001

%F G.f.: Sum_{n>=1} A000005(n^2)*x^n/(1-x^n). - _Mircea Merca_, Feb 25 2014

%F a(n) = A066446(n) + A184389(n). - _Reinhard Zumkeller_, Sep 08 2015

%F Let b(n), n > 0, be the Dirichlet inverse of a(n). Then b(n) is multiplicative with b(p^e) = (-1)^e*(Sum_{i=0..e} binomial(3,i)) for prime p and e >= 0, where binomial(n,k)=0 if n < k; abs(b(n)) is multiplicative and has the Dirichlet g.f.: (zeta(s))^4/(zeta(2*s))^3. - _Werner Schulte_, Feb 07 2021

%p A035116 := proc(n) numtheory[tau](n)^2 ; end proc:

%p seq(A035116(n),n=1..40) ; # _R. J. Mathar_, Apr 02 2011

%t DivisorSigma[0, Range[100]]^2 (* _Vladimir Joseph Stephan Orlovsky_, Jul 20 2011 *)

%o (Magma) [ NumberOfDivisors(n)^2 : n in [1..100] ];

%o (PARI) A035116(n)=numdiv(n)^2;

%o (Haskell)

%o a035116 = (^ 2) . a000005' -- _Reinhard Zumkeller_, Sep 08 2015

%Y Cf. A000005, A048691, A061391.

%Y Cf. A066446, A184389, A061502.

%K nonn,easy,mult

%O 1,2

%A _N. J. A. Sloane_

%E Additional comments from _Vladeta Jovovic_, Apr 29 2001