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A047851 a(n) = A047848(3,n).

%I A047851 #36 May 02 2025 12:54:51
%S A047851 1,2,8,44,260,1556,9332,55988,335924,2015540,12093236,72559412,
%T A047851 435356468,2612138804,15672832820,94036996916,564221981492,
%U A047851 3385331888948,20311991333684,121871948002100,731231688012596,4387390128075572,26324340768453428,157946044610720564,947676267664323380
%N A047851 a(n) = A047848(3,n).
%C A047851 n-th difference of a(n), a(n-1), ..., a(0) is A000351(n-1) for n >= 1.
%H A047851 G. C. Greubel, <a href="/A047851/b047851.txt">Table of n, a(n) for n = 0..1000</a>
%H A047851 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-6).
%F A047851 a(n) = (6^n + 4)/5. - _Ralf Stephan_, Feb 14 2004
%F A047851 From _Philippe Deléham_, Oct 05 2009: (Start)
%F A047851 a(0) = 1, a(1) = 2, a(n) = 7*a(n-1) - 6*a(n-2) for n > 1.
%F A047851 G.f.: (1 - 5*x)/(1 - 7*x + 6*x^2). (End)
%F A047851 a(n) = 6*a(n-1) - 4 (with a(0)=1). - _Vincenzo Librandi_, Aug 06 2010
%F A047851 E.g.f.: exp(x)*(exp(5*x) + 4)/5. - _Elmo R. Oliveira_, Aug 29 2024
%p A047851 a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=6*a[n-1]+1 od: seq(a[n]+1, n=0..20); # _Zerinvary Lajos_, Mar 20 2008
%t A047851 (6^Range[0,40] +4)/5 (* _G. C. Greubel_, Jan 11 2025 *)
%o A047851 (Magma) [(6^n + 4)/5: n in [0..40]]; // _G. C. Greubel_, Jan 11 2025
%o A047851 (Python)
%o A047851 def A047851(n): return (pow(6,n) + 4)//5
%o A047851 print([A047851(n) for n in range(41)]) # _G. C. Greubel_, Jan 11 2025
%Y A047851 Cf. A000351, A047848.
%K A047851 nonn,easy
%O A047851 0,2
%A A047851 _Clark Kimberling_
%E A047851 a(21)-a(24) from _Elmo R. Oliveira_, Aug 29 2024