A106610 Numerator of n/(n+9).

0, 1, 2, 1, 4, 5, 2, 7, 8, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 19, 20, 7, 22, 23, 8, 25, 26, 3, 28, 29, 10, 31, 32, 11, 34, 35, 4, 37, 38, 13, 40, 41, 14, 43, 44, 5, 46, 47, 16, 49, 50, 17, 52, 53, 6, 55, 56, 19, 58, 59, 20, 61, 62, 7, 64, 65, 22, 67, 68, 23, 70, 71, 8, 73, 74, 25, 76, 77

Offset: 0

Author

N. J. A. Sloane, May 15 2005

Keywords

• nonn
• easy
• frac
• mult

Comments

Apart from 0, also numerator of Sum_{i=1..n} (1/((i+2)*(i+3))) = n/(3n+9). - Bruno Berselli, Nov 07 2012

In addition to being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019

Examples

			For n = 12, n/(n+9) = 12/21 = 4/7. So, a(12) = 4. - _Indranil Ghosh_, Jan 31 2017
From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - (2*3)*G(x^3) - (2*9)*G(x^9) , where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (2/3)*H(x^3) - (2/9)*H(x^9), where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (2/3^2)*L(x^3) - (2/9^2)*L(x^9), where L(x) = Log(1/(1 - x)).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (2/3)*L(x^3) + (2/3)*L(x^9). (End)
		

References

• Raffaello Giusti, editore, Supplemento al Periodico di Matematica (Livorno), Jul 1902, p. 138 (Problem 421, case k=3).

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..20000 (terms 0..1000 from Vincenzo Librandi)
• Peter Bala, A note on the sequence of numerators of a rational function, 2019.
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0, 0,-1).

Crossrefs

Cf. A008292, A109050.

Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Programs

GAP

List([0..80],n->NumeratorRat(n/(n+9))); # Muniru A Asiru, Feb 19 2019

Magma

[Numerator(n/(n+9)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

Maple

seq(numer(n/(n+9)),n=0..80); # Muniru A Asiru, Feb 19 2019

Mathematica

f[n_]:=Numerator[n/(n+9)]; Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)

PARI

vector(100, n, n--; numerator(n/(n+9))) \\ G. C. Greubel, Feb 19 2019

Sage

[lcm(n,9)/9for n in range(0, 100)] # Zerinvary Lajos, Jun 09 2009

Formula

From R. J. Mathar, Apr 18 2011: (Start)

a(n) = A109050(n)/9.

Dirichlet g.f. zeta(s-1)*(1-2/3^s-2/9^s).

Multiplicative with a(3^e) = 3^max(0,e-2), a(p^e) = p^e if p<>3. (End)

a(n) = 2*a(n-9) - a(n-18). - G. C. Greubel, Feb 19 2019

From Peter Bala, Feb 21 2019: (Start)

a(n) = n/gcd(n,9), where gcd(n,9) = [1, 1, 3, 1, 1, 3, 1, 1, 9, ...] is a periodic sequence of period 9: a(n) is thus quasi_polynomial in n.

O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - 2*F(x^3) - 2*F(x^9), where F(x) = x/(1 - x)^2.

More generally, for m >= 1, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 3^m)*( F(m,x^3) + F(m,x^9) ), 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 to the o.g.f. for the sequence produces generating functions for the sequences n^m*a(n), m in Z. Some examples are given below. (End)

Sum_{k=1..n} a(k) ~ (61/162) * n^2. - Amiram Eldar, Nov 25 2022