A035116 a(n) = tau(n)^2, where tau(n) = A000005(n).
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, 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, 4, 64, 16, 64, 16, 16, 4, 144, 4, 16
Offset: 1
References
- G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 59.
- G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, Theorem 304.
Links
- T. D. Noe, Table of n, a(n) for n = 1..1000
- Mircea Merca, The Lambert series factorization theorem, The Ramanujan Journal, January 2017; DOI: 10.1007/s11139-016-9856-3.
Programs
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Haskell
a035116 = (^ 2) . a000005' -- Reinhard Zumkeller, Sep 08 2015
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Magma
[ NumberOfDivisors(n)^2 : n in [1..100] ];
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Maple
A035116 := proc(n) numtheory[tau](n)^2 ; end proc: seq(A035116(n),n=1..40) ; # R. J. Mathar, Apr 02 2011
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Mathematica
DivisorSigma[0, Range[100]]^2 (* Vladimir Joseph Stephan Orlovsky, Jul 20 2011 *)
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PARI
A035116(n)=numdiv(n)^2;
Formula
Dirichlet g.f.: zeta(s)^4/zeta(2s).
tau(n)^2 = Sum_{d|n} tau(d^2), Dirichlet convolution of A048691 and A000012 (i.e.: inverse Mobius transform of A048691).
Multiplicative with a(p^e) = (e+1)^2. - Vladeta Jovovic, Dec 03 2001
G.f.: Sum_{n>=1} A000005(n^2)*x^n/(1-x^n). - Mircea Merca, Feb 25 2014
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
Extensions
Additional comments from Vladeta Jovovic, Apr 29 2001