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a(0) = 19; for n>0, successively subtract 5, subtract 3 and double.
1

%I #25 Sep 08 2022 08:45:18

%S 19,14,11,22,17,14,28,23,20,40,35,32,64,59,56,112,107,104,208,203,200,

%T 400,395,392,784,779,776,1552,1547,1544,3088,3083,3080,6160,6155,6152,

%U 12304,12299,12296,24592,24587,24584,49168,49163,49160,98320,98315,98312,196624

%N a(0) = 19; for n>0, successively subtract 5, subtract 3 and double.

%C Suggested by a test found on the Internet.

%H G. C. Greubel, <a href="/A106706/b106706.txt">Table of n, a(n) for n = 0..1000</a>

%H Jeffrey N. Shaumeyer, Bearcastle Blog, <a href="https://web.archive.org/web/20071009070537/http://bearcastle.com/blog/?p=596">One Post, Two ...</a> [Via Internet Archive Wayback-Machine]

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-2).

%F G.f.: (19 + 14*x + 11*x^2 - 35*x^3 - 25*x^4 - 19*x^5)/((1 - 2*x^3)*(1 - x^3)).

%F a(3n) = 3*2^n+16, a(3n+1) = 3*2^n+11, a(3n+2) = 3*2^n+8.

%p f:=proc(n) option remember; if n=0 then RETURN(19); fi; if n mod 3 = 1 then RETURN(f(n-1)-5); elif n mod 3 = 2 then RETURN(f(n-1)-3); else RETURN(2*f(n-1)); fi; end;

%t nxt[{b_,c_,d_}]:={d-5,d-8,2d-16}; Join[{19},Flatten[NestList[nxt,{14,11,22},20]]] (* _Harvey P. Dale_, Dec 01 2019 *)

%o (PARI) a(n)=3*2^(n\3)+[16,11,8][n%3+1] \\ _M. F. Hasler_, Nov 16 2010

%o (Magma) I:=[19,14,11,22,17,14]; [n le 6 select I[n] else 3*Self(n-3) - 2*Self(n-6): n in [1..61]]; // _G. C. Greubel_, Sep 09 2021

%o (Sage)

%o def p(n): return 0 if (n%3==0) else 5 if (n%3==1) else 8

%o def a(n,b): return 2^(n//3)*(b-16) + 16 - p(n)

%o [a(n,19) for n in (0..60)] # _G. C. Greubel_, Sep 09 2021

%K nonn

%O 0,1

%A _N. J. A. Sloane_, based on correspondence with Jeffrey N. Shaumeyer, Apr 23 2006