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Revision History for A135994

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Showing entries 1-10 | older changes
First differences of A135992.
(history; published version)
#19 by Peter Luschny at Tue Sep 27 18:36:55 EDT 2016
STATUS

proposed

approved

#18 by Joerg Arndt at Sun Sep 25 05:53:36 EDT 2016
STATUS

editing

proposed

Discussion
Sun Sep 25
10:29
Michel Marcus: Can you have a look at A154714
#17 by Joerg Arndt at Sun Sep 25 05:53:24 EDT 2016
FORMULA

From Vladimir Reshetnikov, Sep 24 2016 : (Start):

STATUS

proposed

editing

Discussion
Sun Sep 25
05:53
Joerg Arndt: Fixed attribution, please note.
#16 by Vaclav Kotesovec at Sun Sep 25 02:14:49 EDT 2016
STATUS

editing

proposed

#15 by Vaclav Kotesovec at Sun Sep 25 02:14:33 EDT 2016
FORMULA

From Vladimir Reshetnikov, Sep 24 2016 (BeginStart):

STATUS

proposed

editing

#14 by Wesley Ivan Hurt at Sat Sep 24 20:10:08 EDT 2016
STATUS

editing

proposed

#13 by Wesley Ivan Hurt at Sat Sep 24 20:09:54 EDT 2016
LINKS

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-1).

FORMULA

a(n) = 3*a(n-2) - a(n-4) for n>3. G.f.: -x*(x-2) / ((x^2-x-1)*(x^2+x-1)). [Colin Barker, Feb 02 2013]

MATHEMATICA

Table[(LucasL[n] - (-1)^n Fibonacci[n + 3])/2, {n, 0, 2040}] (* Vladimir Reshetnikov, Sep 24 2016 *)

STATUS

proposed

editing

#12 by Vladimir Reshetnikov at Sat Sep 24 18:59:32 EDT 2016
STATUS

editing

proposed

#11 by Vladimir Reshetnikov at Sat Sep 24 18:59:10 EDT 2016
FORMULA

a(n) = Sum_{k=01..n} (-1)^(k+1) * Fibonacci(k) * Lucas(n-k).

a(n) = (LucasLLucas(n) - (-1)^n * Fibonacci(n+3))/2, where Fibonacci(n) = A000045(n), Lucas(n) = A000032(n). (End)

CROSSREFS
#10 by Vladimir Reshetnikov at Sat Sep 24 18:56:36 EDT 2016
FORMULA

From Vladimir Reshetnikov, Sep 24 2016 (Begin):

a(n) = Sum_{k=0..n} (-1)^(k+1) * Fibonacci(k) * Lucas(n-k).

a(n) = (LucasL(n) - (-1)^n * Fibonacci(n+3))/2. (End)

MATHEMATICA

Table[(LucasL[n] - (-1)^n Fibonacci[n + 3])/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 24 2016 *)

STATUS

approved

editing