|
|
A067745
|
|
Numerator of ((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))).
|
|
11
|
|
|
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 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
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
Numerator of n/2^n + (n-1)/2^(n-1), two Oresme numbers. - Paul Curtz, Dec 07 2021
|
|
LINKS
|
|
|
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
|
|
MAPLE
|
f:= n -> (3*n-2)/2^padic:-ordp(3*n-2, 2):
|
|
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
|
|
|
KEYWORD
|
easy,frac,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|