A079291 - OEIS

%I #108 Dec 03 2024 11:19:04

%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 (-1)^(n+1)*a(n) is the r=-4 member of the r-" 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

%C In general, squaring the terms of a Horadam sequence with signature (c,d) will result in a third-order recurrence with signature (c^2+d, c^2*d+d^2, -d^3). - _Gary Detlefs_, Nov 11 2021

%C (Conjectured) For any primitive Pythagorean triple of the form (X, Y, Z=Y+1), it appears that Y or Z will always be (and only be) a square Pell number if X = A001333(n), for n > 1. If n is even, Y is always a square Pell number, and if n is odd, then Z is always a square Pell number. For example: (3, 4, 5), (7, 24, 25), (17, 144, 145), (41, 840, 841), (99, 4900, 4901). - _Jules Beauchamp_, Feb 02 2022

%C a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, black half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), and white half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using black (1/4,1/4)-fences, white (1/4,1/4)-fences, and (1/4,3/4)-fences. - _Michael A. Allen_, Dec 29 2022

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

%H Michael A. Allen and Kenneth Edwards, <a href="https://www.fq.math.ca/Papers1/60-5/allen.pdf">Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.

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

%H Sergio Falcon, <a href="https://ikprress.org/index.php/AJOMCOR/article/view/442">Some series of reciprocal k-Fibonacci numbers</a>, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; <a href="https://www.researchgate.net/publication/297715665_SOME_SERIES_OF_RECIPROCAL_k-FIBONACCI_NUMBERS">ResearchGate link</a>.

%H Toufik Mansour, <a href="https://arxiv.org/abs/math/0302015">A note on sum of k-th power of Horadam's sequence</a>, arXiv:math/0302015 [math.CO], 2003.

%H Toufik Mansour, <a href="https://arxiv.org/abs/math/0303138">Squaring the terms of an ell-th order linear recurrence</a>, arXiv:math/0303138 [math.CO], 2003.

%H Pantelimon Stanica, <a href="https://arxiv.org/abs/math/0010149">Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences</a>, arXiv:math/0010149 [math.CO], 2000.

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

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

%F G.f.: x*(1-x)/((1+x)*(1-6*x+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) is the rightmost term of M^n * [1 0 0] where M is 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 Lim_{k -> infinity} ( a(n+k)/a(k) ) = A001541(n) + 2*A001109(n)*sqrt(2). - _Johannes W. Meijer_, Aug 01 2010

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

%F a(n) = (1/8)*(A002203(2*n) - 2*(-1)^n). - _G. C. Greubel_, Sep 17 2021

%F Conjectured formula for (X, Y, Z) for primitive Pythagorean triple of the form (X, Y, Z=Y+1) is (A001333(n)^2, A079291(n)^2, A079291(n)^2-1) or (A001333(n)^2), A079291(n)^2-1, A079291(n)^2). As a closed formula (X, Y, Z) =  ((1-sqrt(2))^n + (1+sqrt(2)) ^n))/2, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2)- 4)/8, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2)+4)/8. - _Jules Beauchamp_, Feb 02 2022

%F From _Michael A. Allen_, Dec 29 2022: (Start)

%F a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.

%F a(n+1) = (1 + (-1)^n)/2 + 4*Sum_{k=1..n} ( k*a(n+1-k) ). (End)

%F Product_{n>=2} (1 + (-1)^n/a(n)) = (1 + sqrt(2))/2 (A174968) (Falcon, 2016, p. 189, eq. (3.1)). - _Amiram Eldar_, Dec 03 2024

%p with(combinat):seq(fibonacci(i,2)^2, i=0..31); # _Zerinvary Lajos_, Mar 20 2008

%t CoefficientList[Series[x(1-x)/((1+x)*(1-6x+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 *)

%t Fibonacci[Range[0, 30], 2]^2 (* _G. C. Greubel_, Sep 17 2021 *)

%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..31]]; // _Vincenzo Librandi_, May 17 2013

%o (Sage) [lucas_number1(n, 2, -1)^2 for n in (0..30)] # _G. C. Greubel_, Sep 17 2021

%Y Cf. A000129, A001254, A001109, A001541, A001653, A174968.

%Y Cf. A002203, A007598, A084158 (partial sums), A086346, A092184.

%K easy,nonn

%O 0,3

%A _Ralf Stephan_, Feb 08 2003