A181318 a(n) = A060819(n)^2.

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

Offset: 0

Author

Paul Curtz, Jan 26 2011

Keywords

• nonn
• mult
• easy

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)

a(n-4) is the constant needed to complete the n-polygonal numbers into squares (see A377851); a(-1) = 1, which completes the triangle numbers, is not shown in the data. - Jonathan Dushoff, Nov 12 2024

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000
• Wikipedia, Completing the square.
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Crossrefs

Cf. A181829, A046970, A007434, A060819, A168077, A377851.

Programs

Magma

[n^2/GCD(n,4)^2: n in [0..100]]; // G. C. Greubel, Sep 19 2018

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 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,1,1,9,1,25,9,49,4,81,25,121},60] (* Harvey P. Dale, Jan 18 2025 *)

PARI

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

Sage

[n^2/gcd(n, 4)^2 for n in (0..100)] # G. C. Greubel, Feb 20 2019

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)

Sum_{n>=1} 1/a(n) = 5*Pi^2/12. - Amiram Eldar, Aug 12 2022

From Amiram Eldar, Nov 25 2022: (Start)

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)

a(n) = (37 - 27*(-1)^n - 3*(-1)^(n*(n-1)/2) - 3*(-1)^(n*(n+1)/2)) * n^2/64. - Vaclav Kotesovec, Nov 14 2024

Extensions

Edited by Jean-François Alcover, Oct 01 2012 and Jan 15 2013

More terms from Michel Marcus, Jun 09 2014