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A079291
Squares of Pell numbers.
27
0, 1, 4, 25, 144, 841, 4900, 28561, 166464, 970225, 5654884, 32959081, 192099600, 1119638521, 6525731524, 38034750625, 221682772224, 1292061882721, 7530688524100, 43892069261881, 255821727047184, 1491038293021225
OFFSET
0,3
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
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.
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
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 *)
PROG
(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
(Sage) [lucas_number1(n, 2, -1)^2 for n in (0..30)] # G. C. Greubel, Sep 17 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Ralf Stephan, Feb 08 2003
STATUS
approved