login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A170931
Extended Lucas L(n,i) = n*(L(n,i-1) + L(n,i-2)) = a^i + b^i where d = sqrt(n*(n+4)); a=(n+d)/2; b=(n-d)/2.
5
2, 4, 24, 112, 544, 2624, 12672, 61184, 295424, 1426432, 6887424, 33255424, 160571392, 775307264, 3743514624, 18075287552, 87275208704, 421401985024, 2034708774912, 9824443039744, 47436607258624, 229044201193472
OFFSET
0,1
COMMENTS
Sequence gives the rational part of the radii of the circles in nested circles and squares inspired by Vitruvian Man, starting with a square whose sides are of length 4 (in some units). The radius of the circle is an integer in the real quadratic number field Q(sqrt(2)), namely R(n) = A(n-1) + B(n)*sqrt(2) with A(-1)=1, for n >= 1, A(n-1) = A170931(n-1)*-1^(n-1); and B(n) = A094013(n)*-1^n. See illustrations in the links. - Kival Ngaokrajang, Feb 15 2015
LINKS
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
FORMULA
From R. J. Mathar, Feb 05 2010: (Start)
a(n) = 2*A084128(n) = 4*a(n-1) + 4*a(n-2).
G.f.: 2*(1-2*x)/(1 - 4*x - 4*x^2). (End)
EXAMPLE
L(n,0)=2, L(n,1)=n.
MATHEMATICA
CoefficientList[Series[2 (1 - 2 x) / (1 - 4 x - 4 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 16 2015 *)
LinearRecurrence[{4, 4}, {2, 4}, 30] (* Harvey P. Dale, Sep 03 2016 *)
PROG
(PARI) x='x+O('x^30); Vec(2*(1-2*x)/(1 - 4*x - 4*x^2)) \\ G. C. Greubel, Dec 21 2017
(Magma) I:=[2, 4]; [n le 5 select I[n] else 4*Self(n-1)+4*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017
CROSSREFS
Cf. similar sequences with d=sqrt(n*(n+k)): A000032 (k=1, classic Lucas), A080040 (k=2), A085480 (k=3).
Sequence in context: A119036 A192382 A232205 * A317999 A371892 A164313
KEYWORD
nonn,easy
AUTHOR
Claudio Peruzzi (claudio.peruzzi(AT)gmail.com), Feb 04 2010
STATUS
approved