Search: id:A060789 Results 1-1 of 1 results found. %I A060789 %S A060789 1,1,1,2,5,1,7,4,3,5,11,2,13,7,5,8,17,3,19,10,7,11,23,4,25,13,9,14,29, %T A060789 5,31,16,11,17,35,6,37,19,13,20,41,7,43,22,15,23,47,8,49,25,17,26,53,9, %U A060789 55,28,19,29,59,10,61,31,21,32,65,11,67,34,23,35,71,12,73,37,25,38,77 %N A060789 a(n) = n / (gcd(n,2) * gcd(n,3)). %C A060789 a(n+2) is absolute value of numerator of determinant of n X n matrix with M(i,j) = 2/(i(i+1)) if i=j otherwise 1. - Alexander Adamchuk (alex(AT)kolmogorov.com), May 19 2006 %H A060789 Harry J. Smith, Table of n, a(n) for n=1,...,1000 %F A060789 G.f.: (x + x^2 + x^3 + 2x^4 + 5x^5 + x^6 + 5x^7 + 2x^8 + x^9 + x^10 + x^11)/(x^6 - 1)^2. %F A060789 Multiplicative with a(2^e)=2^(e-1), a(3^e)=3^(e-1), a(p^e)=p^e, p>3. - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 09 2004 %F A060789 a(n) = Numerator[(-1)^(n+1)*Det[DiagonalMatrix[Table[2/(i(i+1))-1, {i, 1,n-2}]]+1]], n>2. - Alexander Adamchuk (alex(AT)kolmogorov.com), May 19 2006 %F A060789 a(n) divides n. a(6k) = k for integer k>0. a(p^k) = p^k for prime p>3 and integerk>0. - Alexander Adamchuk (alex(AT)kolmogorov.com), Sep 20 2006 %t A060789 Numerator[Table[(-1)^(n+1) Det[ DiagonalMatrix[ Table[ 2/(i(i+1)) - 1, {i, 1, n-2} ] ] + 1 ], {n, 1, 30} ]] - Alexander Adamchuk (alex(AT)kolmogorov.com), May 19 2006 %t A060789 Table[Numerator[(n+3)/(n+2)/(n+1)/n],{n,60}] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 17 2009] %o A060789 (Other) sage: [lcm(n,6)/6for n in xrange(1, 78)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 07 2009] %o A060789 (PARI) { for (n=1, 1000, write("b060789.txt", n, " ", n / (gcd(n, 2) * gcd(n, 3))); ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jul 11 2009] %Y A060789 Sequence in context: A065224 A165278 A106619 this_sequence A134570 A019510 A124576 %Y A060789 Adjacent sequences: A060786 A060787 A060788 this_sequence A060790 A060791 A060792 %K A060789 nonn,easy,mult %O A060789 1,4 %A A060789 Len Smiley (smiley(AT)math.uaa.alaska.edu), Apr 26 2001 %E A060789 More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001 Search completed in 0.001 seconds