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%I #30 Mar 06 2024 15:45:24
%S 3,10,22,43,78,136,231,386,638,1047,1710,2784,4523,7338,11894,19267,
%T 31198,50504,81743,132290,214078,346415,560542,907008,1467603,2374666,
%U 3842326,6217051,10059438,16276552,26336055,42612674,68948798,111561543,180510414
%N Convolution of natural numbers >= 3 and (Fib(2), Fib(3), Fib(4), ...).
%C a(n) is the sum of row n in the triangle T(n,k) defined by: T(n,1) = T(n,n) = 2*n+1 for n>=1 and T(n,k) = 3*T(n-1,k-1) - 2*T(n-1,k) + T(n-2,k-1) for n>2, 2<=k<=n-1. - _Lechoslaw Ratajczak_, Nov 07 2020
%C Floretion Algebra Multiplication Program, FAMP code: (a(n)) = 4jesleftforcycseq[ - .25'i + .5'k - .25i' - .5j' + .5k' - .75'ii' + .75'jj' - .25'kk' + .25'jk' - .5'ki' + .25'kj' + .25e ], apart from initial terms. 4jesrightforcycseq = A022308; 2jesforcycseq(n+2) = n+2; identity: jesleft + jesright = jes; vesforcycseq was set to the constant sequence = (-1,-1,-1,-1,-1...). (Dement)
%H Colin Barker, <a href="/A023554/b023554.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).
%F G.f.: x*(1+x)*(3-2*x) / ((1-x)^2*(1-x-x^2)).
%F 2*(n+5) = A022308(n+4) - a(n+1) (conjectured). Note offset of A022308 is 0. - _Creighton Dement_, Feb 02 2005
%F From _Colin Barker_, Feb 20 2017: (Start)
%F a(n) = -7 + (2^(-1-n)*((1-t)^n*(-19+9*t) + (1+t)^n*(19+9*t)))/t - 2*(1+n) where t=sqrt(5).
%F a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>4. (End)
%F a(n) = Fibonacci(n+5) + 2*Fibonacci(n+3) - (2*n + 9). - _G. C. Greubel_, Jul 08 2019
%F a(n) = a(n-1) + a(n-2) + 2*n + 3 for n>2. - _Lechoslaw Ratajczak_, Nov 07 2020
%t Table[Fibonacci[n+5] + 2*Fibonacci[n+3] -2*n-9, {n, 40}] (* _G. C. Greubel_, Jul 08 2019 *)
%o (PARI) Vec(x*(1+x)*(3-2*x) / ((1-x)^2*(1-x-x^2)) + O(x^60)) \\ _Colin Barker_, Feb 20 2017
%o (PARI) vector(40, n, f=fibonacci; f(n+5)+2*f(n+3)-(2*n+9)) \\ _G. C. Greubel_, Jul 08 2019
%o (Magma) F:=Fibonacci; [F(n+5)+2*F(n+3)-(2*n+9): n in [1..40]]; // _G. C. Greubel_, Jul 08 2019
%o (SageMath) f=fibonacci; [f(n+5)+2*f(n+3)-(2*n+9) for n in (1..40)] # _G. C. Greubel_, Jul 08 2019
%o (GAP) F:=Fibonacci;; List([1..40], n-> F(n+5)+2*F(n+3)-(2*n+9)) # _G. C. Greubel_, Jul 08 2019
%Y Cf. A000045, A213584.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_
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