|
%I #39 Mar 26 2024 04:09:38
%S 25,1225,112225,11122225,1111222225,111112222225,11111122222225,
%T 1111111222222225,111111112222222225,11111111122222222225,
%U 1111111111222222222225,111111111112222222222225,11111111111122222222222225,1111111111111222222222222225,111111111111112222222222222225
%N a(n) = A133473(n+1)^2.
%H Vincenzo Librandi, <a href="/A181719/b181719.txt">Table of n, a(n) for n = 1..100</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).
%F a(n) = 100 * A181718(n-1) + 25.
%F a(n) = 25 * A109344(n-1), for n > 1.
%F From _Colin Barker_, Aug 21 2019: (Start)
%F G.f.: x*(1 - 62*x + 160*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)).
%F a(n) = (5 + 10^n)^2 / 9.
%F a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3. (End)
%F E.g.f.: (1/9)*exp(x)*(25 + 10*exp(9*x) + exp(99*x)). - _Stefano Spezia_, Aug 21 2019 after _Colin Barker_
%t (5+10^Range[30])^2/9 (* _G. C. Greubel_, Mar 25 2024 *)
%o (PARI) a(n)=(100^n+10*10^n+25)/9 \\ _Charles R Greathouse IV_, Jun 01 2011
%o (PARI) Vec(5*x*(1 - 62*x + 160*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^17)) \\ _Colin Barker_, Aug 21 2019
%o (Magma) [(100^n+10*10^n+25)/9: n in [1..20]]; // _Vincenzo Librandi_, Jun 02 2011
%o (SageMath) [(5+10^n)^2//9 for n in range(1,31)] # _G. C. Greubel_, Mar 25 2024
%Y Cf. A109344, A133473, A181718.
%K nonn,easy
%O 1,1
%A _Paul Curtz_, Nov 17 2010
%E Formulas edited by _Eric M. Schmidt_, Oct 29 2012
| 