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%I #10 Dec 29 2023 12:53:45
%S 1,3,8,18,38,89,189,389,898,1898,3898,8989,18989,38989,89898,189898,
%T 389898,898989,1898989,3898989,8989898,18989898,38989898,89898989,
%U 189898989,389898989,898989898,1898989898,3898989898,8989898989,18989898989,38989898989,89898989898
%N a(1)=1, a(2)=3, a(3)=8; for n>=4, a(n) = 10*a(n-3) + 8 (if a(n-3) is odd) or + 9 (if a(n-3) is even).
%C From a puzzle (1,3,8,18,?,89,189) given on a civil service test.
%C Another possibility is that 1,3,8,18,?,89,189,... is an erroneous version of A117727. - _Hugo van der Sanden_, Apr 14 2006
%H G. C. Greubel, <a href="/A117713/b117713.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,9,-9,0,10,-10).
%F From _G. C. Greubel_, Jul 23 2023: (Start)
%F a(n) = (1/198)*(2*(89*b(n) + 188*b(n-1) + 386*b(n-2)) + 6*(A010892(n) + A010892(n-1)) - 187 + 3*(-1)^n), where b(n) = 10^floor(n/3)*floor((n-1 mod 3)/2).
%F G.f.: x*(1 + 2*x + 5*x^2 + x^3 + 2*x^4 + 6*x^5)/((1-x^2)*(1-x+x^2)*(1-10*x^3)). (End)
%p f:=proc(n) option remember; local t1; if n=1 then RETURN(1); fi; if n=2 then RETURN(3); fi; if n=3 then RETURN(8); fi; t1:=10*f(n-3)+8; if f(n-3) mod 2 = 0 then t1:=t1+1; fi; RETURN(t1); end;
%t a[n_]:= a[n]= If[n<4, Fibonacci[2*n], 10*a[n-3] +If[Mod[a[n-3], 2]==1, 8, 9]];
%t Table[a[n], {n, 40}] (* _G. C. Greubel_, Jul 23 2023 *)
%o (Magma) I:=[1,3,8,18,38,89,189]; [n le 7 select I[n] else Self(n-1) +9*Self(n-3) -9*Self(n-4) +10*Self(n-6) -10*Self(n-7): n in [1..40]]; // _G. C. Greubel_, Jul 23 2023
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A117713
%o if (n<4): return fibonacci(2*n)
%o elif (a(n-3)%2)==1: return 10*a(n-3) + 8
%o else: return 10*a(n-3) + 9
%o [a(n) for n in range(1,41)] # _G. C. Greubel_, Jul 23 2023
%Y Cf. A010892, A117727.
%K nonn,easy
%O 1,2
%A Louis Ciotti (lciotti(AT)twcny.rr.com), Apr 13 2006
%E Solution proposed by Mohammed BOUAYOUN (mohammed.bouayoun(AT)yahoo.fr), Apr 14 2006
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