A158623 Denominator of the reduced fraction A158620(n)/A158621(n).
9, 18, 10, 45, 63, 28, 108, 135, 55, 198, 234, 91, 315, 360, 136, 459, 513, 190, 630, 693, 253, 828, 900, 325, 1053, 1134, 406, 1305, 1395, 496, 1584, 1683, 595, 1890, 1998, 703, 2223, 2340, 820, 2583, 2709, 946, 2970, 3105, 1081, 3384, 3528, 1225, 3825
Offset: 2
Examples
a(2) = 9 = denominator of (2^3-1)/2^3+1 = 7/9. a(3) = 18 = denominator of ((2^3-1)*(3^3-1))/((2^3+1)*(3^3+1)) = (7 * 26)/ (9 * 28) = 182/252 = 13/18. a(4) = 10 = denominator of ((2^3-1)*(3^3-1)*(4^3-1))/((2^3+1)*(3^3+1)*(4^3+1)) = (7 * 26 * 63)/(9 * 28 * 65) = 11466/16380 = 7/10. a(5) = 45 = denominator of ((2^3-1)(3^3-1)(4^3-1)(5^3-1))/((2^3+1)(3^3+1)(4^3+1)(5^3+1)) = 1421784/2063880 = 31/45.
Programs
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Maple
A158623 := proc(n) 2*(n^2+n+1)/3/n/(n+1) ; denom(%) ; end: seq(A158623(n),n=2..100) ; # R. J. Mathar, Mar 27 2009
Formula
Denominator of (Product_{k=2..n} (k^3-1)) / Product_{k=2..n} (k^3+1) = denominator of Product_{k=2..n} A068601(k)/A001093(k).
A158620(n)/A158621(n) = 2(n^2+n+1)/(3n(n+1)). Conjecture: a(n) = 3a(n-3) - 3a(n-6) + a(n-9), so trisections are A152996, A060544 and 3*A081266. - R. J. Mathar, Mar 27 2009
Empirical g.f.: -x^2*(x^8 - 2*x^5 + 9*x^4 + 18*x^3 + 10*x^2 + 18*x + 9) / ((x-1)^3*(x^2 + x + 1)^3). - Colin Barker, May 09 2013
Extensions
More terms from R. J. Mathar, Mar 27 2009
Comments