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A001922
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Numbers k such that 3*k^2 - 3*k + 1 is both a square (A000290) and a centered hexagonal number (A003215).
(Formerly M4569 N1946)
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7
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1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481
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OFFSET
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0,2
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COMMENTS
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Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)
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REFERENCES
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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FORMULA
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a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
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EXAMPLE
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8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - Colin Barker, Jan 07 2015
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MAPLE
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seq(simplify((1 +ChebyshevU(n, 7) +ChebyshevU(n-1, 7))/2), n=0..30); # G. C. Greubel, Oct 07 2022
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MATHEMATICA
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With[{s1=3+2Sqrt[3], s2=3-2Sqrt[3], t1=7+4Sqrt[3], t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12, {n, 0, 20}]]] (* or *) LinearRecurrence[ {15, -15, 1}, {1, 8, 105}, 21] (* Harvey P. Dale, Aug 14 2011 *)
CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 16 2012 *)
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PROG
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(Magma) I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012
(PARI) Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015
(SageMath) [(1+chebyshev_U(n, 7) +chebyshev_U(n-1, 7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022
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CROSSREFS
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KEYWORD
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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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