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A055210
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Sum of totients of square divisors of n.
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2
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1, 1, 1, 3, 1, 1, 1, 3, 7, 1, 1, 3, 1, 1, 1, 11, 1, 7, 1, 3, 1, 1, 1, 3, 21, 1, 7, 3, 1, 1, 1, 11, 1, 1, 1, 21, 1, 1, 1, 3, 1, 1, 1, 3, 7, 1, 1, 11, 43, 21, 1, 3, 1, 7, 1, 3, 1, 1, 1, 3, 1, 1, 7, 43, 1, 1, 1, 3, 1, 1, 1, 21, 1, 1, 21, 3, 1, 1, 1, 11, 61, 1, 1, 3, 1, 1, 1, 3, 1, 7, 1, 3, 1, 1, 1, 11, 1
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OFFSET
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1,4
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LINKS
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FORMULA
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a(n) = Sum_{d is square and divides n} phi(d).
Multiplicative with a(p^e) = (p^(e+1)+1)/(p+1) for even e and a(p^e) = (p^e+1)/(p+1) for odd e. - Vladeta Jovovic, Dec 01 2001
G.f.: Sum_{k>=1} k * phi(k) * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Aug 20 2021
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = zeta(3/2)/(3*zeta(2)) = 0.529377... . - Amiram Eldar, Nov 13 2022
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EXAMPLE
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n = 400: its square divisors are {1, 4, 16, 25, 100, 400}, their totients are {1, 2, 8, 20, 40, 160} and the totient-sum over these divisors is, so a(400) = 231. This value arises at special squarefree multiples of 400 (400 times 2, 3, 5, 6, 7, 10, 11, 13, 15, 17, 19, 21, 22, 23 etc).
a(400) = a(2^4*5^2) = (2^5 + 1)/3*(5^3 + 1)/6 = 231.
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MATHEMATICA
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Array[DivisorSum[#, EulerPhi, IntegerQ@ Sqrt@ # &] &, 97] (* Michael De Vlieger, Nov 18 2017 *)
f[p_, e_] := If[EvenQ[e], (p^(e + 1) + 1)/(p + 1), (p^e + 1)/(p + 1)]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 09 2020 *)
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PROG
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(PARI) a(n) = sumdiv(n, d, eulerphi(d)*issquare(d)); \\ Michel Marcus, Dec 31 2013
(Magma) [&+[EulerPhi(d):d in Divisors(n)| IsSquare(d)]: n in [1..100]]; // Marius A. Burtea, Oct 14 2019
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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