OFFSET
2,1
COMMENTS
A158620(n) = Product_{k=2..n} (k^3-1). A158621(n) = Product_{k=2..n} (k^3+1). A158622(n) is the numerator of the reduced fraction A158620(n)/A158621(n). A158623(n) is the denominator of the reduced fraction A158620(n)/A158621(n). The reduced fractions are 7/9, 13/18, 7/10, 31/45, 43/63, 19/28, 73/108, 91/135, 37/55, 133/198, ...
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
EXAMPLE
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.
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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Mar 23 2009
EXTENSIONS
More terms from R. J. Mathar, Mar 27 2009
STATUS
approved