The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A131040

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A131040

a(n) = (1/2+1/2*i*sqrt(11))^n + (1/2-1/2*i*sqrt(11))^n, where i=sqrt(-1).
(history; published version)

#15 by N. J. A. Sloane at Sat Dec 07 12:18:25 EST 2019

PROG

(Sage) [lucas_number2(n, 1, 3) for n in xrangerange(1, 34)] # Zerinvary Lajos, May 14 2009

Discussion

Sat Dec 07

12:18

OEIS Server: https://oeis.org/edit/global/2837

#14 by Jon E. Schoenfield at Thu Dec 24 23:08:28 EST 2015

STATUS

proposed

approved

#13 by Jon E. Schoenfield at Thu Dec 24 18:50:32 EST 2015

STATUS

editing

proposed

#12 by Jon E. Schoenfield at Thu Dec 24 18:50:30 EST 2015

NAME

a(n) = (1/2+1/2*Ii*sqrt(11))^n + (1/2-1/2*Ii*sqrt(11))^n, where Ii=sqrt(-1).

STATUS

approved

editing

#11 by Bruno Berselli at Fri Jun 26 04:47:01 EDT 2015

STATUS

editing

approved

#10 by Bruno Berselli at Fri Jun 26 04:46:57 EDT 2015

NAME

a(n) = (1/2+1/2*I*sqrt(11))^n + (1/2-1/2*I*sqrt(11))^n, where I=sqrt(-1).

STATUS

proposed

editing

#9 by Michel Marcus at Fri Jun 26 04:09:14 EDT 2015

STATUS

editing

proposed

#8 by Michel Marcus at Fri Jun 26 04:09:04 EDT 2015

PROG

(Sage) [lucas_number2(n, 1, 3) for n in xrange(1, 34)] # [From __Zerinvary Lajos_, May 14 2009]

STATUS

proposed

editing

#7 by Peter Bala at Fri Jun 26 04:01:40 EDT 2015

STATUS

editing

proposed

#6 by Peter Bala at Tue Jun 23 15:31:47 EDT 2015

COMMENTS

Essentially the Lucas sequence V(1,3). - Peter Bala, Jun 23 2015

LINKS

Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>

FORMULA

a(n) = a(n-1) - 3*a(n-2); G.f. (1 - 6*x)/(1 - x + 3*x^2+1-x).

a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x - 11*x^2))/2 )^n. - Peter Bala, Jun 23 2015

STATUS

approved

editing