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).

1, 1, 4, 6, 13, 23, 43, 77, 138, 244, 429, 749, 1301, 2249, 3872, 6642, 11357, 19363, 32927, 55861, 94566, 159776, 269469, 453721, 762793, 1280593, 2147068, 3595422, 6013933, 10048559, 16773139, 27971549, 46605186, 77587084, 129063117, 214531397, 356346557

Offset: 1

Author

Gary W. Adamson, Dec 28 2007

Keywords

• nonn
• easy

Comments

By definition, the arithmetic mean of a(1) ... a(n) is equal to A000045(n).

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

Examples

			a(6) = 23 = 6*F(6) - 5*F(5) = 6*8 - 5*5 = 48 - 25.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Crossrefs

Cf. A000045, A246715, A354044.

Programs

Julia

# The function 'fibrec' is defined in A354044.
function A136391(n)
    a, b = fibrec(n - 1)
    n*b - (n - 1)*a
end
println([A136391(n) for n in 1:35]) # Peter Luschny, May 18 2022

Maple

with(combinat): seq(n*fibonacci(n)-(n-1)*fibonacci(n-1),n=1..30); # Emeric Deutsch, Jan 01 2008

Mathematica

Table[n Fibonacci[n] - (n-1) Fibonacci[n-1], {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)

PARI

Vec(x*(1-x)*(1+x^2)/(1-x-x^2)^2 + O(x^100)) \\ Altug Alkan, Oct 28 2015

Formula

Equals A128064 * A000045.

From R. J. Mathar, Nov 25 2008: (Start)

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) = A045925(n) - A045925(n-1).

G.f.: x*(1 - x)*(1 + x^2)/(1 - x - x^2)^2.

a(n) = A014286(n-1) - A014286(n-2), n>3. (End)

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

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

a(n) = F(n-1) + n*F(n-2). - Bruno Berselli, Jul 26 2017

Extensions

More terms from Emeric Deutsch, Jan 01 2008