%I #50 Oct 15 2024 14:37:29
%S 1,3,1,7,9,11,13,3,17,19,21,23,5,27,29,31,33,7,37,39,41,43,9,47,49,51,
%T 53,11,57,59,61,63,13,67,69,71,73,15,77,79,81,83,17,87,89,91,93,19,97,
%U 99,101,103,21,107,109,111,113,23,117,119,121,123,25,127,129,131,133
%N Denominator of (9*(n^4 - 2*n^3 + 2*n^2 - n) + 2)/(2*(2*n - 1)).
%C From _Altug Alkan_, Apr 13 2018: (Start)
%C Also numerator of (2*n-1)/5.
%C Proof: Since 9*(n^4-2*n^3+2*n^2-n)+2 = 9*n^4-18*n^3+18*n^2-9*n+2 = (3*n^2-3*n+1)*(3*n^2-3*n+2), this is an even number. So we can see that 9*n^4-18*n^3+18*n^2-9*n+2 = (4*n^3-7*n^2+5*n-2)*(2*n-1) + n^2*(n^2+1) and we should focus on the n^2*(n^2+1)/(2*n-1) part for the denominator, n^2+1 = ((2*n-1)/4)*((2*n+1)+5/(2*n-1)) and n^2*(n^2+1)/(2*n-1) = (n^2/4)*(2*n+1+5/(2*n-1)).
%C Since gcd(n^2, 2*n-1) = 1 and 4 is always killed by the numerator part independent of denominator of 5/(2*n-1), the denominator of (9*(n^4-2*n^3+2*n^2-n)+2)/(2*(2*n-1)) will always be determined by the denominator of 5/(2*n-1).
%C In other words, this is the numerator of (2*n-1)/5. (End)
%H G. C. Greubel, <a href="/A096431/b096431.txt">Table of n, a(n) for n = 1..2000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MagicHexagon.html">Magic Hexagon</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,2,0,0,0,0,-1).
%F Satisfies a linear recurrence with characteristic polynomial (1-x^5)^2.
%F G.f.: x*(1+x)*(1+2*x-x^2+8*x^3+x^4+8*x^5-x^6+2*x^7+x^8)/((1-x)^2*(1+x+x^2+x^3+x^4)^2). - _R. J. Mathar_, Mar 11 2011
%e 1, 28/3, 38, 703/7, 1891/9, 4186/11, ...
%t Table[ (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)),{n,80}]//Denominator (* or *)
%t LinearRecurrence[{0,0,0,0,2,0,0,0,0,-1},{1,3,1,7,9,11,13,3,17,19},80] (* _Harvey P. Dale_, Aug 25 2021 *)
%o (PARI) a(n) = numerator((2*n-1)/5); \\ _Altug Alkan_, Apr 13 2018
%o (PARI) first(n) = my(res = vector(n, i, 2*i - 1)); forstep(i = 3, n, 5, res[i]/=5); res \\ _David A. Corneth_, Apr 15 2018
%o (Magma)
%o A096431:= func< n | Numerator((2*n-1)/5) >; [A096431(n): n in [1..60]]; // _G. C. Greubel_, Oct 14 2024
%o (SageMath)
%o def A096431(n): return numerator((2*n-1)/5)
%o [A096431(n) for n in range(1,61)] # _G. C. Greubel_, Oct 14 2024
%Y Cf. A096430 (numerators), A097361, A146535.
%K nonn,easy,frac
%O 1,2
%A _Eric W. Weisstein_, Aug 09 2004