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A136391 a(n) = n*F(n) - (n-1)*F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

%I A136391 #39 Jan 30 2025 05:22:46
%S A136391 1,1,4,6,13,23,43,77,138,244,429,749,1301,2249,3872,6642,11357,19363,
%T A136391 32927,55861,94566,159776,269469,453721,762793,1280593,2147068,
%U A136391 3595422,6013933,10048559,16773139,27971549,46605186,77587084,129063117,214531397,356346557
%N A136391 a(n) = n*F(n) - (n-1)*F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).
%C A136391 By definition, the arithmetic mean of a(1) ... a(n) is equal to A000045(n).
%C A136391 Proof of the three-term recurrence formula: a(n+1) - a(n) - a(n-1) = ((n+1)*F(n+1) - n*F(n)) - (n*F(n) - (n-1)*F(n-1)) - ((n-1)*F(n-1) - (n-2)*F(n-2)) = (n+1)*F(n+1) - 2*n*F(n) + (n-2)*F(n-2) = (n+1)*(2*F(n) - F(n-2)) - 2*n*F(n) + (n-2)*F(n-2) = 2*F(n) - 3*F(n-2) = F(n-1) + F(n-3) = L(n-2). - _Giuseppe Coppoletta_, Sep 01 2014
%H A136391 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-2,-1).
%F A136391 Equals A128064 * A000045.
%F A136391 From _R. J. Mathar_, Nov 25 2008: (Start)
%F A136391 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) = A045925(n) - A045925(n-1).
%F A136391 G.f.: x*(1 - x)*(1 + x^2)/(1 - x - x^2)^2.
%F A136391 a(n) = A014286(n-1) - A014286(n-2), n>3. (End)
%F A136391 Recurrence: a(n+1) = a(n) + a(n-1) + L(n-2) for n>1, where L = A000032 (see proof in Comments section). - _Giuseppe Coppoletta_, Sep 01 2014
%F A136391 E.g.f.: (exp(x*phi)/phi+exp(-x/phi)*phi)*(x+1)/sqrt(5)-1, where phi=(1+sqrt(5))/2. - _Vladimir Reshetnikov_, Oct 28 2015
%F A136391 a(n) = F(n-1) + n*F(n-2). - _Bruno Berselli_, Jul 26 2017
%e A136391 a(6) = 23 = 6*F(6) - 5*F(5) = 6*8 - 5*5 = 48 - 25.
%p A136391 with(combinat): seq(n*fibonacci(n)-(n-1)*fibonacci(n-1),n=1..30); # _Emeric Deutsch_, Jan 01 2008
%t A136391 Table[n Fibonacci[n] - (n-1) Fibonacci[n-1], {n, 1, 20}] (* _Vladimir Reshetnikov_, Oct 28 2015 *)
%o A136391 (PARI) Vec(x*(1-x)*(1+x^2)/(1-x-x^2)^2 + O(x^100)) \\ _Altug Alkan_, Oct 28 2015
%o A136391 (Julia) # The function 'fibrec' is defined in A354044.
%o A136391 function A136391(n)
%o A136391     a, b = fibrec(n - 1)
%o A136391     n*b - (n - 1)*a
%o A136391 end
%o A136391 println([A136391(n) for n in 1:35]) # _Peter Luschny_, May 18 2022
%Y A136391 Cf. A000045, A246715, A354044.
%K A136391 nonn,easy
%O A136391 1,3
%A A136391 _Gary W. Adamson_, Dec 28 2007
%E A136391 More terms from _Emeric Deutsch_, Jan 01 2008