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A071954
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a(n) = 4*a(n-1) - a(n-2) - 4, with a(0) = 2, a(1) = 4.
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8
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2, 4, 10, 32, 114, 420, 1562, 5824, 21730, 81092, 302634, 1129440, 4215122, 15731044, 58709050, 219105152, 817711554, 3051741060, 11389252682, 42505269664, 158631825970, 592022034212, 2209456310874, 8245803209280, 30773756526242, 114849222895684
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OFFSET
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0,1
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COMMENTS
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a(n) gives the side of a cube having a square number of cubes in its two outermost layers, i.e., solutions p to the equation p^3 - (p - 4)^3 = q^2. The corresponding q is given by 4*A001075(n).
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REFERENCES
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M. E. Larsen, "Four Cubes" in Puzzler's Tribute, Ed. D. Wolfe & T. Rodgers, pp. 69-70, A. K. Peters, MA, 2002
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LINKS
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FORMULA
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a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 2, with a(0) = 2, a(1) = 4, a(2) = 10.
G.f.: 2*(1 - 3*x)/((1-x)*(1 -4*x +x^2)). - Harvey P. Dale, May 05 2011
a(n) = (2 + (-(2 - sqrt(3))^n + (2 + sqrt(3))^n)/sqrt(3)). - Colin Barker, Nov 03 2016
a(n) = 2*(1 - (-i)^(n+1)*F(n, 4*i)), where i=sqrt(-1) and F(n,x) is the Fibonacci polynomial. (End)
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EXAMPLE
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G.f. = 2 + 4*x + 10*x^2 + 32*x^3 + 114*x^4 + 420*x^5 + 1562*x^6 + ...
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MATHEMATICA
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a[n_]:= a[n] = 4*a[n-1] -a[n-2] -4; a[0]=2; a[1]=4; Table[a[n], {n, 0, 30}]
LinearRecurrence[{5, -5, 1}, {2, 4, 10}, 30] (* Harvey P. Dale, May 05 2011 *)
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PROG
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(Haskell)
a071954 n = a071954_list !! n
a071954_list = 2 : 4 : zipWith (-)
(map ((4 *) . pred) (tail a071954_list)) a071954_list
(PARI) {a(n) = my(w=quadgen(12)); simplify( 2 + ((2+w)^n - (2-w)^n) / w)}; /* Michael Somos, Nov 03 2016 */
(Magma) I:=[2, 4, 10]; [n le 3 select I[n] else 5*Self(n-1) -5*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 25 2019
(Sage) (2*(1-3*x)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
(GAP) a:=[2, 4, 10];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019
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CROSSREFS
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KEYWORD
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nice,nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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