A096431 - OEIS

A096431

Denominator of (9*(n^4 - 2*n^3 + 2*n^2 - n) + 2)/(2*(2*n - 1)).

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, 53, 11, 57, 59, 61, 63, 13, 67, 69, 71, 73, 15, 77, 79, 81, 83, 17, 87, 89, 91, 93, 19, 97, 99, 101, 103, 21, 107, 109, 111, 113, 23, 117, 119, 121, 123, 25, 127, 129, 131, 133

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OFFSET

1,2

COMMENTS

From Altug Alkan, Apr 13 2018: (Start)

Also numerator of (2*n-1)/5.

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)).

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).

In other words, this is the numerator of (2*n-1)/5. (End)

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..2000

Eric Weisstein's World of Mathematics, Magic Hexagon

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).

FORMULA

Satisfies a linear recurrence with characteristic polynomial (1-x^5)^2.

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

EXAMPLE

1, 28/3, 38, 703/7, 1891/9, 4186/11, ...

MATHEMATICA

Table[ (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)), {n, 80}]//Denominator (* or *)