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A125301 a(3n) = n, a(3n+1) = (n+2)*a(3n), a(3n+2) = (n+2)*a(3n+1). 1

%I #10 Jul 31 2018 15:50:02

%S 0,0,0,1,3,9,2,8,32,3,15,75,4,24,144,5,35,245,6,48,384,7,63,567,8,80,

%T 800,9,99,1089,10,120,1440,11,143,1859,12,168,2352,13,195,2925,14,224,

%U 3584,15,255,4335,16,288,5184,17,323,6137,18,360,7200,19,399,8379,20

%N a(3n) = n, a(3n+1) = (n+2)*a(3n), a(3n+2) = (n+2)*a(3n+1).

%C From a competency test.

%H Colin Barker, <a href="/A125301/b125301.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,4,0,0,-6,0,0,4,0,0,-1).

%F From _Colin Barker_, Jan 24 2017: (Start)

%F a(n) = 4*a(n-3) - 6*a(n-6) + 4*a(n-9) - a(n-12) for n>11.

%F G.f.: x^3*(1 + 3*x + 9*x^2 - 2*x^3 - 4*x^4 - 4*x^5 + x^6 + x^7 + x^8) / ((1 - x)^4*(1 + x + x^2)^4).

%F (End)

%p a:=proc(n): if n mod 3=0 then n/3 elif n mod 3 = 1 then (n+5)*a(n-1)/3 else (n+4)*a(n-1)/3 fi end: seq(a(n),n=0..75); # _Emeric Deutsch_, Jan 01 2007

%t LinearRecurrence[{0,0,4,0,0,-6,0,0,4,0,0,-1},{0,0,0,1,3,9,2,8,32,3,15,75},70] (* _Harvey P. Dale_, Jul 31 2018 *)

%o (PARI) concat(vector(3), Vec(x^3*(1 + 3*x + 9*x^2 - 2*x^3 - 4*x^4 - 4*x^5 + x^6 + x^7 + x^8) / ((1 - x)^4*(1 + x + x^2)^4) + O(x^100))) \\ _Colin Barker_, Jan 24 2017

%K nonn,easy

%O 0,5

%A Trent Stratton (trent.stratton(AT)gov.bc.ca), Dec 08 2006

%E Solution from _N. J. A. Sloane_, Dec 08 2006

%E More terms from _Emeric Deutsch_, Jan 01 2007

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Last modified April 19 15:11 EDT 2024. Contains 371794 sequences. (Running on oeis4.)