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A181318 a(n) = A060819(n)^2. 13
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

The first sequence, p=0, of the family A060819(n)*A060819(n+p).

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.

From Peter Bala, Feb 19 2019: (Start)

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)

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

FORMULA

a(2*n) = A168077(n), a(2*n+1) = A016754(n).

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

From Peter Bala, Feb 19 2019: (Start)

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)

MAPLE

a:=n->n^2/gcd(n, 4)^2: seq(a(n), n=0..60); # Muniru A Asiru, Feb 20 2019

MATHEMATICA

Table[n^2/GCD[n, 4]^2, {n, 0, 100}] (* G. C. Greubel, Sep 19 2018 *)

PROG

(PARI) a(n)=n^2/gcd(n, 4)^2 \\ Charles R Greathouse IV, Dec 21 2011

(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

CROSSREFS

Cf. A181829, A046970, A007434, A060819, A168077.

Sequence in context: A205381 A237587 A191871 * A202006 A195278 A092477

Adjacent sequences:  A181315 A181316 A181317 * A181319 A181320 A181321

KEYWORD

nonn,mult,easy,changed

AUTHOR

Paul Curtz, Jan 26 2011

EXTENSIONS

Edited by Jean-Fran├žois Alcover, Oct 01 2012 and Jan 15 2013

More terms from Michel Marcus, Jun 09 2014

STATUS

approved

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Last modified March 18 13:47 EDT 2019. Contains 321289 sequences. (Running on oeis4.)