The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A264080

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A264080

a(n) = 6*F(n)*F(n+1) + (-1)^n, where F = A000045.
(history; published version)

#48 by Charles R Greathouse IV at Thu Sep 08 08:46:14 EDT 2022

PROG

(MAGMAMagma) [6*Fibonacci(n)*Fibonacci(n+1)+(-1)^n: n in [0..30]];

Discussion

Thu Sep 08

08:46

OEIS Server: https://oeis.org/edit/global/2944

#47 by N. J. A. Sloane at Tue Jan 07 23:47:39 EST 2020

STATUS

editing

approved

#46 by N. J. A. Sloane at Tue Jan 07 23:47:11 EST 2020

COMMENTS

Let (A), (B) and (C) be sequences of the Fibonacci type, x = {1, 2, ...}, y = {0, 1, ...}. Form the three sequences with the initial values A(0) = x, A(1) = y, B(0) = 7*x - y, B(1) = 5*x, C(0) = A(n), C(1) = B(n). Then a(n) = C(n)/x always applies. - Klaus Purath, Oct 28 2019

FORMULA

From Klaus Purath, Oct 28 2019: (Start)

(a(n-3) - a(n-2) - a(n-1) + a(n))/6 = Fibonacci(2*n-2).

(a(n-5) + a(n))/30 = Fibonacci(2*n-4).

(a(n) - a(n-4))/18 = Fibonacci(2*n-3). (End)

STATUS

reviewed

editing

Discussion

Tue Jan 07

23:47

N. J. A. Sloane: I rejected the proposed formulas

#45 by G. C. Greubel at Mon Dec 30 14:56:55 EST 2019

STATUS

proposed

reviewed

#44 by Klaus Purath at Mon Dec 16 08:00:17 EST 2019

STATUS

editing

proposed

#43 by Klaus Purath at Mon Dec 16 07:57:47 EST 2019

COMMENTS

Let (A), (B) and (C) be sequences of the Fibonacci type, x = {1, 2, ...}, y = {0, 1, ...}. Form the three sequences with the initial values A(0) = x, A(1) = y, B(0) = 7*x - y, B(1) = 115*x, C(0) = A(n), C(1) = B(n). Then a(n) = C(n)/x always applies. - Klaus Purath, Oct 28 2019

STATUS

proposed

editing

Discussion

Mon Dec 16

08:00

Klaus Purath: Typo in comment corrected.

#42 by Stefano Spezia at Mon Dec 09 11:23:31 EST 2019

STATUS

editing

proposed

#41 by Stefano Spezia at Mon Dec 09 11:23:17 EST 2019

FORMULA

E.g.f.: (1/5)*exp(-x)*(-1 + 6*exp(5*x/2)*(cosh((sqrt(5)*x)/2) + sqrt(5)*sinh((sqrt(5)*x)/2))). - Stefano Spezia, Dec 09 2019

STATUS

proposed

editing

#40 by Klaus Purath at Tue Nov 19 17:25:31 EST 2019

STATUS

editing

proposed

#39 by Klaus Purath at Tue Nov 19 17:24:38 EST 2019

FORMULA

a(n) = (a(n-2) - a(n-1) - a(n+1) + a(n+2))/4.