# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a087954 Showing 1-1 of 1 %I A087954 #12 Oct 11 2017 05:08:10 %S A087954 40,320,2160,15004,103680,709804,4868640,33385280,228811000, %T A087954 1568358004,10749957120,73681030804,505018447960,3461452808000, %U A087954 23725145626560,162614587921804,1114577054219520,7639424691459004,52361396168994000,358890350005878080 %N A087954 (2 + phi)/a(n) is the sum of successive remainders when computing the Euclidean algorithm for (1, A088166(n)/phi) with phi being the golden ratio, where n >= 2. %H A087954 Colin Barker, Table of n, a(n) for n = 2..1000 %F A087954 For n = 3*k, a(n) = Lucas(4*n) - 2. For all other n, a(n) = Lucas(4*n) - Lucas(2*n). %F A087954 Empirical g.f.: 4*x^2*(10 + 10*x - 10*x^2 - 139*x^3 + 13*x^4 - 48*x^5 + 144*x^6 - 16*x^7 + 2*x^8 - 7*x^9 + x^10) / ((1 - x)*(1 - 7*x + x^2)*(1 - 3*x + x^2)*(1 + x + x^2)*(1 + 3*x + 8*x^2 + 3*x^3 + x^4)). - _Colin Barker_, Mar 10 2016 %o A087954 (PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1) %o A087954 a(n) = if(n%3==0, lucas(4*n)-2, lucas(4*n)-lucas(2*n)) \\ _Colin Barker_, Mar 10 2016 %Y A087954 Cf. A088166. %K A087954 easy,nonn %O A087954 2,1 %A A087954 _Thomas Baruchel_, Sep 21 2003 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE