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A097988
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a(n) = Sum_{d dividing n} tau(d)^3 = (Sum_{d dividing n} tau(d))^2.
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5
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1, 9, 9, 36, 9, 81, 9, 100, 36, 81, 9, 324, 9, 81, 81, 225, 9, 324, 9, 324, 81, 81, 9, 900, 36, 81, 100, 324, 9, 729, 9, 441, 81, 81, 81, 1296, 9, 81, 81, 900, 9, 729, 9, 324, 324, 81, 9, 2025, 36, 324, 81, 324, 9, 900, 81, 900, 81, 81, 9, 2916, 9, 81, 324
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OFFSET
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1,2
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COMMENTS
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When n = p^e is a prime power, we have the corollary a(n) = Sum_{r=1..e+1} r^3 = (Sum_{r=1..e+1} r)^2, i.e. A000537(n) = (A000217(n))^2.
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REFERENCES
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Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 47.
Jean-Marie De Koninck and Armel Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 562, pp. 75, 265; Ellipses Paris 2004.
William J. LeVeque, Topics in Number Theory. Addison-Wesley, Reading, MA, 2 vols., 1956, Vol. 1, p. 85, Problem 2.
William J. LeVeque, Fundamentals of Number Theory, Dover Publications Inc, 1977, p. 125.
Joe Roberts, The Lure of Integers, MAA, 1992, Integer 3, pages 8-9.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 84.
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LINKS
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FORMULA
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a(n) = (Sum_{d dividing n} (tau(d))^2 = (A007425(n))^2.
Multiplicative with a(p^e) = ((e+1)*(e+2)/2)^2. - Amiram Eldar, Sep 20 2020
Dirichlet g.f.: zeta(s)^5 * Product_{p prime} (1 + 4/p^s + 1/p^(2*s)). - Amiram Eldar, Sep 14 2023
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MAPLE
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with(numtheory); f:=proc(n) local t1; t1:=divisors(n); add(sigma[0](i), i in t1)^2; end;
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MATHEMATICA
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tau[1, n_Integer] := 1; SetAttributes[tau, Listable]; tau[k_Integer, n_Integer] := Plus@@(tau[k-1, Divisors[n]]); A097988[n_] := tau[3, n]^2; Table[A097988[n], {n, 100}] (* Enrique Pérez Herrero, Jul 12 2010 *)
f[n_]:=Total[DivisorSigma[0, Divisors[n]]]^2; f/@Range[100] (* Ivan N. Ianakiev, Mar 05 2015 *)
f[p_, e_] := ((e+1)*(e+2)/2)^2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 20 2020 *)
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PROG
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(PARI) a(n)=sumdiv(n, d, numdiv(d)^3); \\ Michel Marcus, Nov 21 2013
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CROSSREFS
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KEYWORD
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nonn,mult,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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