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0, 1, 1, 9, 1, 25, 9, 49, 4, 81, 25, 121, 9, 169, 49, 225, 16, 289, 81, 361, 25, 441, 121, 529, 36, 625, 169, 729, 49, 841, 225, 961, 64, 1089, 289, 1225, 81, 1369, 361, 1521, 100, 1681, 441, 1849, 121, 2025, 529, 2209, 144, 2401, 625, 2601, 169, 2809, 729
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OFFSET
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0,4
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COMMENTS
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Hence array
p=0: 0, 1, 1, 9, 1, 25, 9, 49, a(n)=A060819(n)^2,
p=1: 0, 1, 3, 3, 5, 15, 21, 14, A064038(n),
p=2: 0, 3, 1, 15, 3, 35, 6, 63, A198148(n),
p=3: 0, 1, 5, 9, 7, 10, 27, 35, A160050(n),
p=4: 0, 5, 3, 21, 2, 45, 15, 77, A061037(n),
p=5: 0, 3, 7, 6, 9, 25, 33, 21, A178242(n),
p=6: 0, 7, 2, 27, 5, 55, 9, 91, A217366(n),
p=7: 0, 2, 9, 15, 11, 15, 39, 49, A217367(n),
p=8: 0, 9, 5, 33, 3, 65, 21, 105, A180082(n).
Compare columns 2, 3 and 5, columns 4 and 7 and columns 6 and 8.
We make some general remarks about the sequence a(n) = numerator(n^2/(n^2 + k^2)) = (n/gcd(n,k))^2 for k a fixed positive integer (we suppress the dependence of a(n) on k). The present sequence corresponds to the case k = 4.
a(n) is a quasi-polynomial in n. In fact, a(n) = n^2/b(n) where b(n) = gcd(n^2,k^2) is a purely periodic sequence in n.
In addition to being multiplicative these sequences are also strong divisibility sequences, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, it follows that a(n) is a divisibility sequence: if n divides m then a(n) divides a(m).
By the multiplicativeness and strong divisibility property of the sequence a(n) it follows that if gcd(n,m) = 1 then a( a(n)*a(m) ) = a(a(n)) * a(a(m)), a( a(a(n))*a(a(m)) ) = a(a(a(n))) * a(a(a(m))) and so on.
The sequence a(n) has the rational generating function Sum_{d divides k} f(d)*F(x^d), where F(x) = x*(1 + x)/(1 - x)^3 = x + 4*x^2 + 9*x^3 + 16*x^4 + ... is the o.g.f. for the squares A000290, and where f(n) is the Dirichlet inverse of the Jordan totient function J_2(n) - see A007434. The function f(n) is multiplicative and is defined on prime powers p^k by f(p^k) = (1 - p^2). See A046970. Cf. A060819. (End)
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
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FORMULA
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a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
G.f.: x*(1 + x + 9*x^2 + x^3 + 22*x^4 + 6*x^5 + 22*x^6 + x^7 + 9*x^8 + x^9 + x^10)/(1-x^4)^3. - R. J. Mathar, Mar 10 2011
a(n) = numerator(n^2/(n^2 + 16)) = n^2/(gcd(n^2,16)) = (n/gcd(n,4))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 16, 1, 4, 1, 16, ...] is a purely periodic sequence of period 4.
a(n) is a quasi-polynomial in n: a(4*n) = n^2; a(4*n + 1) = (4*n + 1)^2; a(4*n + 2) = (2*n + 1)^2; a(4*n + 3) = (4*n + 3)^2.
O.g.f.: Sum_{d divides 4} A046970(d)*x^d*(1 + x^d)/(1 - x^d)^3 = x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3 - 3*x^4*(1 + x^4)/(1 - x^4)^3. (End)
Multiplicative with a(2^e) = 4^max(0, e-2), and a(p^e) = p^(2*e) for p > 2.
Dirichlet g.f.: zeta(s-2)*(1 - 3/2^s - 3/4^s).
Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)
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MAPLE
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a:=n->n^2/gcd(n, 4)^2: seq(a(n), n=0..60); # Muniru A Asiru, Feb 20 2019
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MATHEMATICA
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Table[n^2/GCD[n, 4]^2, {n, 0, 100}] (* G. C. Greubel, Sep 19 2018 *)
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PROG
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(Magma) [n^2/GCD(n, 4)^2: n in [0..100]]; // G. C. Greubel, Sep 19 2018
(Sage) [n^2/gcd(n, 4)^2 for n in (0..100)] # G. C. Greubel, Feb 20 2019
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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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