cp's OEIS Frontend

A051109 Expansion of g.f. (1+2*x+5*x^2)/(1-10*x^3).

%I A051109 #29 Jul 27 2023 12:25:08
%S A051109 1,2,5,10,20,50,100,200,500,1000,2000,5000,10000,20000,50000,100000,
%T A051109 200000,500000,1000000,2000000,5000000,10000000,20000000,50000000,
%U A051109 100000000,200000000,500000000,1000000000,2000000000,5000000000,10000000000,20000000000,50000000000
%N A051109 Expansion of g.f. (1+2*x+5*x^2)/(1-10*x^3).
%H A051109 G. C. Greubel, <a href="/A051109/b051109.txt">Table of n, a(n) for n = 0..1000</a>
%H A051109 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,10).
%F A051109 a(3*n) = 10^n, a(3*n+1) = 2*10^n, a(3*n+2) = 5*10^n.
%F A051109 a(n) = ( 1 + (n mod 3)^2 )*10^floor(n/3). - Justin L. Brown (jlbrown(AT)neo.tamu.edu), Jun 17 2004
%F A051109 G.f.: (1+2*x+5*x^2)/(1-10*x^3). - _Philippe Deléham_, Apr 08 2013
%F A051109 a(n) = 10*a(n-3) with n>2, a(0)=1, a(1)=2, a(2)=5. - _Philippe Deléham_, Apr 08 2013
%F A051109 From _Amiram Eldar_, Jul 27 2023: (Start)
%F A051109 Sum_{n>=0} 1/a(n) = 17/9.
%F A051109 Sum_{n>=0} (-1)^n/a(n) = 7/11. (End)
%t A051109 a[n_]:= a[n]= If[n<3, Fibonacci[2n+1], 10*a[n-3]];
%t A051109 Table[a[n], {n,0,40}] (* _G. C. Greubel_, Jul 23 2023 *)
%o A051109 (Python) print( [ ((n % 3) ** 2 + 1) * 10**int(n/3) for n in range(100)] )
%o A051109 (Magma) [(1 +(n mod 3)^2)*10^Floor(n/3): n in [0..40]]; // _G. C. Greubel_, Jul 23 2023
%o A051109 (SageMath) [(1 +(n%3)^2)*10^(n//3) for n in range(41)] # _G. C. Greubel_, Jul 23 2023
%Y A051109 Cf. A117727.
%K A051109 easy,nonn
%O A051109 0,2
%A A051109 Robert Lozyniak (11(AT)onna.com)
%E A051109 Second formula corrected by _Peter C. Lauterbach_, Nov 12 2010
%E A051109 New name using g.f. from _Joerg Arndt_, Jul 23 2023