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a(n) = 2*a(n-1) + 14*a(n-2) for n>1, a(0)=1, a(1)=1.
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%I #26 Jan 01 2024 11:44:38

%S 1,1,16,46,316,1276,6976,31816,161296,768016,3794176,18340576,

%T 89799616,436367296,2129929216,10369000576,50557010176,246280028416,

%U 1200358199296,5848636796416,28502288382976,138885491915776

%N a(n) = 2*a(n-1) + 14*a(n-2) for n>1, a(0)=1, a(1)=1.

%C Binomial transform of A001024 (powers of 15), with interpolated zeros.

%C a(n) is the number of compositions of n when there are 1 type of 1 and 15 types of other natural numbers. - _Milan Janjic_, Aug 13 2010

%H G. C. Greubel, <a href="/A133345/b133345.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,14).

%F G.f.: (1-x)/(1-2*x-14*x^2).

%F a(n) = Sum_{k=0..n} A098158(n,k)*15^(n-k). - _Philippe Deléham_, Dec 26 2007

%F If p[1]=1, and p[i]=15, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - _Milan Janjic_, Apr 29 2010

%F a(n) = (b*i)^(n-1)*(b*i*ChebyshevU(n, -i/b) - ChebyshevU(n-1, -i/b)), with b = sqrt(14). - _G. C. Greubel_, Oct 15 2022

%t LinearRecurrence[{2,14},{1,1},30] (* _Harvey P. Dale_, Jan 07 2016 *)

%o (PARI) Vec((1-x)/(1-2*x-14*x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 12 2012

%o (Magma) [n le 2 select 1 else 2*(Self(n-1) +7*Self(n-2)): n in [1..41]]; // _G. C. Greubel_, Oct 15 2022

%o (SageMath)

%o A133345=BinaryRecurrenceSequence(2,14,1,1)

%o [A133345(n) for n in range(41)] # _G. C. Greubel_, Oct 15 2022

%Y Cf. A001024, A098158.

%K nonn,easy

%O 0,3

%A _Philippe Deléham_, Dec 21 2007