OFFSET
1,3
COMMENTS
From Peter Bala, Feb 19 2019: (Start)
We make some general remarks about the sequence a(n) = numerator(n/(n + k)) = n/gcd(n,k) for k a fixed positive integer. The present sequence is the case k = 4. Several other cases are listed in the Crossrefs. 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.
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..1000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).
FORMULA
G.f.: x*(1 +x +3*x^2 +x^3 +3*x^4 +x^5 +x^6)/(1 - x^4)^2.
a(n) = 2*a(n-4) - a(n-8).
a(n) = (n/16)*(11 - 5*(-1)^n - i^n - (-i)^n). - Ralf Stephan, Mar 15 2003
a(2*n+1) = a(4*n+2) = 2*n+1, a(4*n+4) = n+1. - Ralf Stephan, Jun 10 2005
Multiplicative with a(2^e) = 2^max(0, e-2), a(p^e) = p^e, p >= 3. - Mitch Harris, Jun 29 2005
a(n) = A167192(n+4,4). - Reinhard Zumkeller, Oct 30 2009
From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109045(n)/4.
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/2^(2s)). (End)
a(n+4) - a(n) = A176895(n). - Paul Curtz, Apr 05 2011
a(n) = numerator(Sum_{k=1..n} 1/((k+1)*(k+2))). This summation has a closed form of 1/2 - 1/(n+2) and denominator of A145979(n). - Gary Detlefs, Sep 16 2011
a((2*n-1)*2^p) = ceiling(2^(p-2))*(2*n-1), p >= 0 and n >= 1. - Johannes W. Meijer, Feb 06 2013
a(n) = n / A109008(n). - Reinhard Zumkeller, Nov 25 2013
a(n) = denominator((2n-4)/n). - Wesley Ivan Hurt, Dec 22 2016
From Peter Bala, Feb 21 2019: (Start)
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4), where F(x) = x/(1 - x)^2.
More generally, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 2^m)*( F(m,x^2) + F(m,x^4) ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence a(n) produces generating functions for the sequences ((n^m)*a(n))n>=1 for m in Z. Some examples are given below.
(End)
Sum_{k=1..n} a(k) ~ (11/32) * n^2. - Amiram Eldar, Nov 25 2022
E.g.f.: x*(8*cosh(x) + sin(x) + 3*sinh(x))/8. - Stefano Spezia, Dec 02 2023
EXAMPLE
From Peter Bala, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - 2*G(x^2) - 4*G(x^4), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (1/2)*H(x^2) - (1/4)*H(x^4), where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4), where L(x) = Log(1/(1 - x)).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4). (End)
MAPLE
MATHEMATICA
f[n_]:= n/GCD[n, 4]; Array[f, 80]
PROG
(Sage) [lcm(n, 4)/4 for n in (1..80)] # Zerinvary Lajos, Jun 07 2009
(PARI) a(n) = { n / gcd(n, 4) } \\ Harry J. Smith, Jul 12 2009
(Haskell)
a060819 n = n `div` a109008 n -- Reinhard Zumkeller, Nov 25 2013
(Magma) [n/GCD(n, 4): n in [1..80]]; // G. C. Greubel, Sep 19 2018
(GAP) List([1..80], n->n/Gcd(n, 4)); # Muniru A Asiru, Feb 20 2019
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
Len Smiley, Apr 30 2001
STATUS
approved