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A096431
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Denominator of (9*(n^4 - 2*n^3 + 2*n^2 - n) + 2)/(2*(2*n - 1)).
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2
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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
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1,2
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COMMENTS
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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)
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).
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FORMULA
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Satisfies a linear recurrence with characteristic polynomial (1-x^5)^2.
G.f.: x*(1+x) *(x^8+2*x^7-x^6+8*x^5+x^4+8*x^3-x^2+2*x+1) / ( (x-1)^2 *(x^4+x^3+x^2+x+1)^2 ). - R. J. Mathar, Mar 11 2011
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EXAMPLE
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1, 28/3, 38, 703/7, 1891/9, 4186/11, ...
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MATHEMATICA
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Table[ (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)), {n, 80}]//Denominator (* or *) 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 *)
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PROG
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(PARI) a(n) = numerator((2*n-1)/5); \\ Altug Alkan, Apr 13 2018
(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
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CROSSREFS
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KEYWORD
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nonn,easy,frac
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AUTHOR
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STATUS
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approved
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