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 A067745 Numerator of ((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))). 7
 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, 19, 79, 41, 85, 11, 91, 47, 97, 25, 103, 53, 109, 7, 115, 59, 121, 31, 127, 65, 133, 17, 139, 71, 145, 37, 151, 77, 157, 5, 163, 83, 169, 43, 175, 89, 181, 23, 187, 95, 193, 49, 199 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Conjecture: Odd part of 3n-2. - Ralf Stephan, Nov 18 2010 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 This sequence is equivalent to A165355 where each element is reduced by the highest possible power of two. - Joe Slater, Nov 30 2016 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 D. S. Mitrinovic and Slavko Simic, Special Functions from Long Ago: 5626, Amer. Math. Monthly 109, (2002), p. 83-84. FORMULA Assuming the above conjecture, a(n) = a((8+(3*n-2)*4^k)/12), for all k >= 1. - L. Edson Jeffery, Feb 15 2015 a(n) = A000265(A165355(n-1)). - Joe Slater, Nov 30 2016 MAPLE f:= n -> (3*n-2)/2^padic:-ordp(3*n-2, 2): map(f, [\$1..100]); # Robert Israel, May 16 2017 MATHEMATICA (* Assuming the above conjecture: *) a067745[n_] := (3*n - 2)/2^IntegerExponent[3*n - 2, 2]; Table[a067745[n], {n, 67}] (* L. Edson Jeffery, Feb 15 2015 *) PROG (PARI) vector(80, n, numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))))) \\ Michel Marcus, Feb 16 2015 (MAGMA) [Numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1)))): n in [1..80]]; // Vincenzo Librandi, Feb 16 2015 CROSSREFS Cf. A067746, A165355. Sequence in context: A247870 A141391 A135766 * A070408 A176672 A107471 Adjacent sequences:  A067742 A067743 A067744 * A067746 A067747 A067748 KEYWORD easy,frac,nonn AUTHOR Marc LeBrun, Jan 29 2002 STATUS approved

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Last modified January 29 08:12 EST 2020. Contains 331337 sequences. (Running on oeis4.)