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Revision History for A099922

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Showing entries 1-10 | older changes
a(n) = F(4n) - 2n, where F(n) = Fibonacci numbers A000045.
(history; published version)
#14 by Alois P. Heinz at Sat May 25 10:27:17 EDT 2019
STATUS

proposed

approved

#13 by Alois P. Heinz at Sat May 25 10:27:10 EDT 2019
STATUS

editing

proposed

#12 by Alois P. Heinz at Sat May 25 10:26:09 EDT 2019
FORMULA

G.f.: x*(1+8x8*x+x^2)/((1-x)^2 * (1-7x7*x+x^2)). [Corrected for offset by Georg Fischer, May 24 2019]

STATUS

reviewed

editing

Discussion
Sat May 25
10:27
Alois P. Heinz: a(0)=0 is default, if nothing else is defined.  So no need to shange data and b-file ...
#11 by Joerg Arndt at Sat May 25 07:44:45 EDT 2019
STATUS

proposed

reviewed

#10 by Michel Marcus at Sat May 25 03:47:32 EDT 2019
STATUS

editing

proposed

Discussion
Sat May 25
04:34
Peter Luschny: I agree with Vaclav. The Fibonacci numbers A000045 are defined for n=0 so it is more reasonable to change the offset here to 0 and prepend a(0)=0. Then the gf is correct, as Vaclav said.
#9 by Michel Marcus at Sat May 25 03:45:58 EDT 2019
FORMULA

a(n) = Sum[_{k=1..n, } Lucas(2n2k-1)^2 ].

STATUS

proposed

editing

#8 by Colin Barker at Sat May 25 03:40:42 EDT 2019
STATUS

editing

proposed

#7 by Colin Barker at Sat May 25 03:39:43 EDT 2019
LINKS

Colin Barker, <a href="/A099922/b099922.txt">Table of n, a(n) for n = 1..1000</a>

FORMULA

From Colin Barker, May 25 2019: (Start)

a(n) = (-((7-3*sqrt(5))/2)^n + ((7+3*sqrt(5))/2)^n)/sqrt(5) - 2*n.

a(n) = 9*a(n-1) - 16*a(n-2) + 9*a(n-3) - a(n-4) for n>4.

(End)

PROG

(PARI) Vec(x*(1 + 8*x + x^2) / ((1 - x)^2*(1 - 7*x + x^2)) + O(x^25)) \\ Colin Barker, May 25 2019

STATUS

proposed

editing

#6 by Vaclav Kotesovec at Sat May 25 02:56:16 EDT 2019
STATUS

editing

proposed

#5 by Vaclav Kotesovec at Sat May 25 02:55:54 EDT 2019
NAME

a(n) = F(4n) - 2n, where F(n) = Fibonacci numbers A000045.