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A079291
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Squares of Pell numbers.
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26
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0, 1, 4, 25, 144, 841, 4900, 28561, 166464, 970225, 5654884, 32959081, 192099600, 1119638521, 6525731524, 38034750625, 221682772224, 1292061882721, 7530688524100, 43892069261881, 255821727047184, 1491038293021225
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OFFSET
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0,3
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COMMENTS
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(-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.
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
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LINKS
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FORMULA
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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) = (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 - sqrt(2))^n + (1 + sqrt(2))^n) /2 )^2 + (-1)^(n+1) )/2. - Antonio Pane (apane1(AT)spc.edu), Dec 15 2007
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
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
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)
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MAPLE
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with(combinat):seq(fibonacci(i, 2)^2, i=0..31); # Zerinvary Lajos, Mar 20 2008
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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