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a(2n) = Lucas(2n+3)^2, a(2n+1) = Lucas(2n+1)^2.
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%I #21 Mar 08 2024 12:15:20

%S 16,1,121,16,841,121,5776,841,39601,5776,271441,39601,1860496,271441,

%T 12752041,1860496,87403801,12752041,599074576,87403801,4106118241,

%U 599074576,28143753121,4106118241,192900153616,28143753121

%N a(2n) = Lucas(2n+3)^2, a(2n+1) = Lucas(2n+1)^2.

%C This sequence is related to several other sequences on squares. In reference to the link "Sequences in Context", (a(n)) = vessigcycseq. Note that the identity "vessigcyc = jessigcyc + lessigcyc + tessigcyc" holds.

%C Floretion Algebra Multiplication Program, FAMP Code: 1vessigcycseq[ + 4.75'i - .75'j - .25'k + 4.75i' - .75j' - .25k' - 1.75'ii' - 3.75'jj' + 4.25'kk' - .25'ij' + 1.25'ik' - .25'ji' + 2.75'jk' + 1.25'ki' + 2.75'kj' + 2.25e]

%H Colin Barker, <a href="/A105671/b105671.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,7,-7,-1,1).

%F G.f.: (-x^4-8x^2+15x-16)/((x-1)(x^4-7x^2+1)).

%F a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n>4. - _Colin Barker_, May 01 2019

%F a(n) = 2*A098149(n+2) +5*A001519(n+2)-2. - _R. J. Mathar_, Sep 11 2019

%t a[n_?EvenQ] := LucasL[n+3]^2; a[n_?OddQ] := LucasL[n]^2; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Sep 28 2011 *)

%o (PARI) Vec((16 - 15*x + 8*x^2 + x^4) / ((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)) + O(x^40)) \\ _Colin Barker_, May 01 2019

%Y Cf. A081071.

%K easy,nonn

%O 0,1

%A _Creighton Dement_, Apr 17 2005