A106617 Numerator of n/(n+16).
0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 2, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 4, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77, 39, 79
Offset: 0
Examples
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) - 8*G(x^8) - 16*G(x^16), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} n^2*a(n)*x^n = H(x) - 2^2*H(x^2) - 4^2*H(x^4) - 8^2*H(x^8) - 16^2*H(x^16), where H(x) = x*(1 + 4*x + x^2)/(1 - x)^4. In general, the o.g.f. for Sum_{n >= 1} (n^k*a(n))*x^n for positive k involves the Eulerian polynomials.
In the other direction,
Sum_{n >= 1} (a(n)/n)*x^n = J(x) - (1/2)*J(x^2) - (1/4)*J(x^4) - (1/8)*J(x^8) - (1/16)*J(x^16), where J(x) = x/(1 - x).
Sum_{n >= 1} (a(n)/n^2)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4) - (1/8^2)*L(x^8) - (1/16^2)*L(x^16), where L(x) = log(1/(1 - x)). In general, the o.g.f. for Sum_{n >= 0} (a(n)/n^k)*x^n, for k >= 3, involves the polylogarithm Li_(k-1)(x).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + (1/2)*L(x^16). (End)
Links
- Nathaniel Johnston, Table of n, a(n) for n = 0..10000
- Peter Bala, A note on the sequence of numerators of a rational function, 2019.
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).
Crossrefs
Programs
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GAP
List([0..100],n->NumeratorRat(n/(n+16))); # Muniru A Asiru, Feb 19 2019
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Magma
[Numerator(n/(n+16)): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
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Maple
seq(numer(n/(n+16)), n=0..100); # Nathaniel Johnston, Apr 18 2011
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Mathematica
Numerator/@Table[n/(n+16),{n,0,10}] (* Harvey P. Dale, Apr 26 2011 *) -
PARI
a(n)=n>>min(valuation(n,2),4) \\ Charles R Greathouse IV, Jun 17 2012
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Sage
[lcm(n,16)/16 for n in range(0, 100)] # Zerinvary Lajos, Jun 12 2009
Formula
a(n) = 2*a(n-16) - a(n-32) for n > 31. - Paul Curtz, Apr 12 2011
Octosections: a(8*n) = A026741(n). a(2+8*n) = 1+4*n. a(4+8*n) = 1+2*n. a(6+8*n) = 3+4*n. Bisection: a(1+2*n) = 1+2*n. - Paul Curtz, Apr 12 2011
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/4^s-1/8^s-1/16^s). - R. J. Mathar, Apr 18 2011
a(n) = numerator of n/(2^(2*n+1)). - Ralf Steiner, Feb 09 2017
The previous comment is incorrect, a(n) first differs from the numerator of n/(2^(2*n+1)) at n = 32. - Peter Bala, Feb 27 2019
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,16), where gcd(n,16) = [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, ...] is a periodic sequence of period 16: a(n) is thus quasi_polynomial in n.
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4) - F(x^8) - F(x^16), 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) + F(m,x^8) + F(m,x^16) ), 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 produces generating functions for the sequences ( (n^m)*a(n) )n>=1 for m in Z. Some examples are given below. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 2^max(0,e-4), and a(p^e) = p^e if p>2.
Sum_{k=1..n} a(k) ~ (171/512) * n^2. (End)
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