OFFSET
1,1
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = 2^(n - A007814(n)).
a(n) = 2*A084623(n). - Paul Curtz, Jan 28 2013
a(n) = 2^A093048(n). - Paul Curtz, Jun 10 2016
From Peter Bala, Feb 25 2019: (Start)
a(n) = 2^n/gcd(n,2^n).
O.g.f.: F(2*x) - (1/2)*F((2*x)^2) - (1/4)*F((2*x)^4) - (1/8)*F((2*x)^8) - ..., where F(x) = x/(1 - x). Cf. A000265.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = F((x/2)) + F((x/2)^2) + 2*F((x/2)^4) + 4*F((x/2)^8) + 8*F((x/2)^16) + 16*F((x/2)^32) + .... (End)
Sum_{n>=1} 1/a(n) = Sum_{n>=1} 2^(2^(n-1)+n-1)/(2^(2^n) - 1) = Sum_{n>=1} A073113(n-1)/A051179(n) = 1.48247501707... - Amiram Eldar, Aug 14 2022
MAPLE
[seq(numer(2^n/n), n=1..50)];
MATHEMATICA
f[n_]:=Numerator[2^n/n]; Array[f, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *)
PROG
(Python)
from fractions import Fraction
def A075101(n):
return (Fraction(2**n)/n).numerator # Chai Wah Wu, Mar 25 2018
(PARI) a(n) = numerator(2^n/n); \\ Michel Marcus, Mar 25 2018
(PARI) a(n) = 2^(n - valuation(n, 2)) \\ Jianing Song, Oct 24 2018
(Magma) [Numerator(2^n/n): n in [1..50]]; // G. C. Greubel, Feb 28 2019
(Sage) [numerator(2^n/n) for n in (1..50)] # G. C. Greubel, Feb 28 2019
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Reinhard Zumkeller, Sep 01 2002
STATUS
approved