A171516 a(n) = a(n-1) + a(n-2) + k, n>1; with a(0) = 1, a(1) = 2, k = 3.

1, 2, 6, 11, 20, 34, 57, 94, 154, 251, 408, 662, 1073, 1738, 2814, 4555, 7372, 11930, 19305, 31238, 50546, 81787, 132336, 214126, 346465, 560594, 907062, 1467659, 2374724, 3842386, 6217113, 10059502, 16276618, 26336123, 42612744, 68948870

Offset: 0

Author

Gary W. Adamson, Dec 10 2009

Keywords

• nonn
• easy

Comments

In an infinite set of sequences converging to phi, H(n+1) = H(n) + H(n-1) + k.

The coincident formula = H(n) = (H(n+1) + H(n-2))/2, then proof of convergence to phi follows, [Gatta and D'Amico]: To get H(4) such that the average of H(4) and H(1) = H(3), the authors write H(4) = 2H(3) - H1 = 2H(1) + 2H(2) + 2k - H(1) = H(2) + (H(1) + H(2) + k) + k = H(2) + H(3) + k, then applying the iterative process to the latter, H(n+1) = H(n) + H(n-1) + k.

Cf. A014739 for a(0) = 1, a(1) = 2, k = 2, getting:

A014739 = (1, 2, 5,. 9, 16, 27, 45, 74, 121, 197,...)

A171516 = (1, 2, 6, 11, 20, 34, 57, 94, 154, 251,...), we obtain

A000071 = (0, 0, 1, .2,..4,..7,.12,.20,..33,..54,...).

Examples

			a(5) = a(4) + a(3) + 3 = 20 + 11 + 3 = 34.
a(4) = (a(5) + a(2)) / 2 = (34 + 6) / 2 = 20.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• F. Gatta and A. D'Amico, Sequences {H(n)} For Which H(n+1)/H(n) Approaches The Golden Ratio, The Fibonacci Quarterly, Vol. 46/47, Nov. 2008/2009, #4.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1). [_R. J. Mathar_, Dec 12 2009]

Crossrefs

Cf. A000045, A014739.

Cf. A000285 (first differences). [From R. J. Mathar, Dec 12 2009]

Programs

GAP

F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1)-3); # G. C. Greubel, Jul 12 2019

Magma

F:=Fibonacci; [F(n+3)+2*F(n+1)-3: n in [0..40]]; // G. C. Greubel, Jul 12 2019

Mathematica

LinearRecurrence[{2,0,-1},{1,2,6},40] (* Harvey P. Dale, Apr 07 2012 *)
With[{F=Fibonacci}, Table[F[n+3]+2*F[n+1]-3, {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)

PARI

vector(40, n, n--; f=fibonacci; f(n+2)+2*f(n+1)-3) \\ G. C. Greubel, Jul 12 2019

Sage

f=fibonacci; [f(n+2)+2*f(n+1)-3 for n in (0..40)] # G. C. Greubel, Jul 12 2019

Formula

a(n) = a(n-1) + a(n-2) + 3, n>1; with a(0) = 1, a(1) = 2.

From R. J. Mathar, Dec 12 2009: (Start)

a(n) = 2*a(n-1) - a(n-3) = A000285(n+1) - 3.

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

a(n) = Fibonacci(n+3) + 2*Fibonacci(n+1) - 3. - G. C. Greubel, Jul 12 2019

Extensions

More terms from R. J. Mathar, Dec 12 2009

Fixed typos in name, formula, crossrefs - Alex Ratushnyak, Apr 27 2012