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%I #44 Jul 19 2024 14:37:09
%S 2,2,8,4,32,32,128,32,512,512,2048,1024,8192,8192,32768,4096,131072,
%T 131072,524288,262144,2097152,2097152,8388608,2097152,33554432,
%U 33554432,134217728,67108864,536870912,536870912,2147483648,134217728,8589934592,8589934592,34359738368,17179869184,137438953472
%N Numerator of 2^n/n.
%H G. C. Greubel, <a href="/A075101/b075101.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 2^(n - A007814(n)).
%F a(n) = 2*A084623(n). - _Paul Curtz_, Jan 28 2013
%F a(n) = 2^A093048(n). - _Paul Curtz_, Jun 10 2016
%F From _Peter Bala_, Feb 25 2019: (Start)
%F a(n) = 2^n/gcd(n,2^n).
%F 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.
%F 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)
%F 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
%p [seq(numer(2^n/n),n=1..50)];
%t f[n_]:=Numerator[2^n/n]; Array[f,50] (* _Vladimir Joseph Stephan Orlovsky_, Feb 16 2011 *)
%o (Python)
%o from fractions import Fraction
%o def A075101(n):
%o return (Fraction(2**n)/n).numerator # _Chai Wah Wu_, Mar 25 2018
%o (PARI) a(n) = numerator(2^n/n); \\ _Michel Marcus_, Mar 25 2018
%o (PARI) a(n) = 2^(n - valuation(n, 2)) \\ _Jianing Song_, Oct 24 2018
%o (Magma) [Numerator(2^n/n): n in [1..50]]; // _G. C. Greubel_, Feb 28 2019
%o (Sage) [numerator(2^n/n) for n in (1..50)] # _G. C. Greubel_, Feb 28 2019
%Y Denominator is A000265(n).
%Y Cf. A007814, A051179, A073113, A084623, A273893.
%K nonn,frac
%O 1,1
%A _Reinhard Zumkeller_, Sep 01 2002
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