A079291 Squares of Pell numbers.

0, 1, 4, 25, 144, 841, 4900, 28561, 166464, 970225, 5654884, 32959081, 192099600, 1119638521, 6525731524, 38034750625, 221682772224, 1292061882721, 7530688524100, 43892069261881, 255821727047184, 1491038293021225

Offset: 0

Author

Ralf Stephan, Feb 08 2003

Keywords

• easy
• nonn

Comments

(-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.

Binomial transform of A086346. - Johannes W. Meijer, Aug 01 2010

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

(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

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
• Joerg Arndt, Matters Computational (The Fxtbook), sect. 32.1.5, pp. 626-627
• Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
• Toufik Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
• Toufik Mansour, Squaring the terms of an ell-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.
• Pantelimon Stanica, Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
• Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

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

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

Programs

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

Maple

with(combinat):seq(fibonacci(i,2)^2, i=0..31); # Zerinvary Lajos, Mar 20 2008

Mathematica

CoefficientList[Series[x(1-x)/((1+x)*(1-6x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, May 17 2013 *)
LinearRecurrence[{5,5,-1},{0,1,4},40] (* Harvey P. Dale, Dec 20 2015 *)
Fibonacci[Range[0, 30], 2]^2 (* G. C. Greubel, Sep 17 2021 *)

Sage

[lucas_number1(n, 2, -1)^2 for n in (0..30)] # G. C. Greubel, Sep 17 2021

Formula

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

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

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

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

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

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

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

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

a(n) = ( (-1)^(n+1) + A001109(n+1) - 3*A001109(n) )/4. - R. J. Mathar, Nov 16 2007

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

Lim_{k -> infinity} ( a(n+k)/a(k) ) = A001541(n) + 2*A001109(n)*sqrt(2). - Johannes W. Meijer, Aug 01 2010

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

a(n) = (1/8)*(A002203(2*n) - 2*(-1)^n). - G. C. Greubel, Sep 17 2021

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

From Michael A. Allen, Dec 29 2022: (Start)

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

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

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