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)

#38 by Alois P. Heinz at Tue Nov 19 14:46:43 EST 2019

STATUS

proposed

editing

Discussion

Tue Nov 19

14:48

Alois P. Heinz: One could add an infitite number of those formulas ... but is is really not helpful.

#37 by Klaus Purath at Mon Oct 28 02:15:40 EDT 2019

STATUS

editing

proposed

Discussion

Sat Nov 16

20:23

Sean A. Irvine: I'm dubious about the utility of this contribution.  You appear to be adding more complicated formulas when there are already plenty of simple explanations.  I wouldn't mind if the new ones made some interesting connections, but these don't.  Also why the extra parentheses in the definition of "(A)" etc. in the comment?

Tue Nov 19

14:46

Alois P. Heinz: The recursions you give do have higher order than needed.  So their value is ... near zero.  Please soo: a recursion or order 3 is given as formula.

#36 by Klaus Purath at Mon Oct 28 02:14:28 EDT 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) = 11*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) = (a(n-2) - a(n-1) - a(n+1) + a(n+2))/4.

(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

approved

editing

#35 by Harvey P. Dale at Fri Jul 12 16:25:51 EDT 2019

STATUS

editing

approved

#34 by Harvey P. Dale at Fri Jul 12 16:25:48 EDT 2019

MATHEMATICA

LinearRecurrence[{2, 2, -1}, {1, 5, 13}, 30] (* Harvey P. Dale, Jul 12 2019 *)

STATUS

approved

editing

#33 by Alois P. Heinz at Wed Sep 28 12:07:06 EDT 2016

STATUS

editing

approved

#32 by Alois P. Heinz at Wed Sep 28 12:06:59 EDT 2016

MAPLE

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1, 5, 13>>)[1, 1]:

seq(a(n), n=0..30); # Alois P. Heinz, Sep 28 2016

STATUS

proposed

editing

#31 by Michel Marcus at Wed Sep 28 11:48:08 EDT 2016

STATUS

editing

proposed

Discussion

Wed Sep 28

11:51

Colin Barker: Merci bien !

#30 by Michel Marcus at Wed Sep 28 11:47:49 EDT 2016

PROG

(PARI) for(n=0, 30, print1(6*fibonacci(n)*fibonacci(n+1)+(-1)^n", "));

STATUS

proposed

editing

Discussion

Wed Sep 28

11:48

Michel Marcus: To have pari scripts together

#29 by Colin Barker at Wed Sep 28 11:43:30 EDT 2016

STATUS

editing

proposed