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A079291 Squares of Pell numbers. 20

%I

%S 0,1,4,25,144,841,4900,28561,166464,970225,5654884,32959081,192099600,

%T 1119638521,6525731524,38034750625,221682772224,1292061882721,

%U 7530688524100,43892069261881,255821727047184,1491038293021225

%N Squares of Pell numbers.

%C a(n)*(-1)^(n+1) is the r=-4 member of the r-family of sequences S_r(n), n>=1, defined in A092184 where more information can be found.

%C Binomial transform of A086346. - _Johannes W. Meijer_, Aug 01 2010

%H Vincenzo Librandi, <a href="/A079291/b079291.txt">Table of n, a(n) for n = 0..1000</a>

%H Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>, sect. 32.1.5, pp. 626-627

%H T. Mansour, <a href="http://arXiv.org/abs/math.CO/0302015">A note on sum of k-th power of Horadam's sequence</a>

%H T. Mansour, <a href="http://arXiv.org/abs/math.CO/0303138">Squaring the terms of an ell-th order linear recurrence</a>

%H P. Stanica, <a href="http://arXiv.org/abs/math.CO/0010149">Generating functions, weighted and non-weighted sums of powers...</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,5,-1).

%F G.f.: x*(1-x)/(1+x)/(1-6x+x^2).

%F a(n) = (r^n+(1/r)^n-2*(-1)^n)/8, with r=3+sqrt(8).

%F a(n+3) = 5*a(n+2)+5*a(n+1)-a(n).

%F L.g.f.: 1/8*log((1+2*x+x^2)/(1-6*x+x^2)) = sum(n>=0, a(n)/n*x^n), see p.627 of the Fxtbook link; special case of the following: let v(0)=0, v(1)=1, and v(n)=u*v(n-1)+v(n-2), then 1/A*log((1+2*x+x^2)/(1-(2-A)*x+x^2)) = sum(n>=0, v(n)^2/n*x^n) where A=u^2+4. [_Joerg Arndt_, Apr 08 2011]

%F a(n+1) = sum_{k=0...n}((-1)^(n-k)*A001653(k)); e.g. 144 = -1 + 5 - 29 + 169; 25 = 1 - 5 + 29 - _Charlie Marion_, Jul 16 2003

%F a(n)=A000129(n)^2.

%F a(n)= (T(n, 3)-(-1)^n)/4 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3)=A001541(n)=((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2. - _Wolfdieter Lang_, Oct 18 2004

%F a(n) = rightmost term of M^n * [1 0 0] where M = the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. a(n+1) = leftmost term. E.g. a(6) = 4900, a(5) = 841 since M^5 * [1 0 0] = [4900 2030 841]. - _Gary W. Adamson_, Oct 31 2004

%F a(n) = [(-1)^(n+1)+A001109(n+1)-3*A001109(n)]/4. - _R. J. Mathar_, Nov 16 2007

%F a(n) = ( ( ( (1-Sqrt[ 2 ])^n + (1+Sqrt[ 2 ])^n) /2 )^2 + (-1)^(n+1) ) /2 - Antonio Pane (apane1(AT)spc.edu), Dec 15 2007

%F Limit(a(n+k)/a(k),k=infinity) = A001541(n)+2*A001109(n)*sqrt(2). - _Johannes W. Meijer_, Aug 01 2010

%F For n>0, a(2n) = 6*a(2n-1)-a(2n-2)-2, a(2n+1) = 6*a(2n)-a(2n-1)+2. - _Charlie Marion_, Sep 24 2011

%p with(combinat):seq(fibonacci(i,2)^2,i=0..21); # _Zerinvary Lajos_, Mar 20 2008

%t CoefficientList[Series[x (1 - x) / (1 + x) / (1 - 6 x + x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, May 17 2013 *)

%t LinearRecurrence[{5,5,-1},{0,1,4},40] (* _Harvey P. Dale_, Dec 20 2015 *)

%o (MAGMA) I:=[0, 1, 4]; [n le 3 select I[n] else 5*Self(n-1)+ 5*Self(n-2) - Self(n-3): n in [1..25]]; // _Vincenzo Librandi_, May 17 2013

%Y Cf. A000129, A001254, A007598, A084158.

%K easy,nonn

%O 0,3

%A _Ralf Stephan_, Feb 08 2003

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Last modified September 26 01:25 EDT 2020. Contains 337346 sequences. (Running on oeis4.)