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A067745 Numerator of ((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))).

%I A067745 #73 Jan 23 2025 15:38:44
%S A067745 1,1,7,5,13,1,19,11,25,7,31,17,37,5,43,23,49,13,55,29,61,1,67,35,73,
%T A067745 19,79,41,85,11,91,47,97,25,103,53,109,7,115,59,121,31,127,65,133,17,
%U A067745 139,71,145,37,151,77,157,5,163,83,169,43,175,89,181,23,187,95,193,49,199
%N A067745 Numerator of ((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))).
%C A067745 Conjecture: Odd part of 3n-2. - _Ralf Stephan_, Nov 18 2010
%C A067745 Conjecture is true.  Note that gcd(3n-2,2n-1)=1 (because 2(3n-2)-3(2n-1) = -1) and gcd(3n-2,n) = 1 or 2.  If 2^k | (3n-2), then k <= log_2(3n-2) < (n-1)/2 for n >= 11.  So only the cases n <= 10 need to be checked individually. - _Robert Israel_, May 16 2017
%C A067745 This sequence is equivalent to A165355 where each element is reduced by the highest possible power of two. - _Joe Slater_, Nov 30 2016
%C A067745 Selecting each odd term gives b(n) = 6n+1 (A016921). A075677 is the even bisection of this sequence, while this sequence is the odd bisection of A075677. - _Cory Kalm_, Apr 29 2021
%C A067745 Numerator of n/2^n + (n-1)/2^(n-1), two Oresme numbers. - _Paul Curtz_, Dec 07 2021
%H A067745 Robert Israel, <a href="/A067745/b067745.txt">Table of n, a(n) for n = 1..10000</a>
%H A067745 D. S. Mitrinovic and Slavko Simic, <a href="http://www.jstor.org/stable/2695786">Special Functions from Long Ago: 5626</a>, Amer. Math. Monthly 109, (2002), p. 83-84.
%F A067745 Assuming the above conjecture, a(n) = a((8+(3*n-2)*4^k)/12), for all k >= 1. - _L. Edson Jeffery_, Feb 15 2015
%F A067745 a(n) = A000265(A165355(n-1)). - _Joe Slater_, Nov 30 2016
%F A067745 a(n) = A000265(3*n-2). - _R. J. Mathar_, Aug 23 2020
%F A067745 a(n) = A075677(2*n-1). a(2*n) = A075677(n); a(2*n-1) = A016921(n). - _Cory Kalm_, May 03 2021
%F A067745 Sum_{k=1..n} a(k) ~ n^2. - _Amiram Eldar_, Aug 26 2024
%F A067745 G.f.: Sum_{k>=1} ((3 + 2*(-1)^(k + 1))*x^(3*2^(k - 1) - (2*(-2)^(k - 1))/3 - 1/3) + (3 - 2*(-1)^(k + 1))*x^(2^(k - 1)*(3 + 2*(-1)^k)/3 - 1/3))/(x^(2^(k + 1)) - 2*x^(2^k) + 1). - _Miles Wilson_, Jan 12 2025
%p A067745 f:= n -> (3*n-2)/2^padic:-ordp(3*n-2,2):
%p A067745 map(f, [$1..100]); # _Robert Israel_, May 16 2017
%t A067745 (* Assuming the above conjecture: *)
%t A067745 a067745[n_] := (3*n - 2)/2^IntegerExponent[3*n - 2, 2]; Table[a067745[n], {n, 67}] (* _L. Edson Jeffery_, Feb 15 2015 *)
%o A067745 (PARI) vector(80, n, numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))))) \\ _Michel Marcus_, Feb 16 2015
%o A067745 (Magma) [Numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1)))): n in [1..80]]; // _Vincenzo Librandi_, Feb 16 2015
%Y A067745 Cf. A067746, A165355, A000265, A075677, A016921, A007310 (range).
%Y A067745 Cf. A075101.
%K A067745 easy,frac,nonn
%O A067745 1,3
%A A067745 _Marc LeBrun_, Jan 29 2002