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A274544
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Values of k such that 2*k-1 and 5*k-1 are both perfect squares.
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2
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1, 13, 925, 18241, 1333345, 26303005, 1922682061, 37928914465, 2772506198113, 54693468355021, 3997952014996381, 78867943439025313, 5765044033118582785, 113727519745606145821, 8313189497804981379085, 163995004605220623248065, 11987613490790750030057281
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = a(n-1) + 1442*a(n-2) - 1442*a(n-3) - a(n-4) + a(n-5) for n>5.
G.f.: x*(1 + 12*x - 530*x^2 + 12*x^3 + x^4) / ((1 - x)*(1 - 38*x + x^2)*(1 + 38*x + x^2)).
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EXAMPLE
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13 is in the sequence because 2*13-1 = 25 = 5^2, and 5*13-1 = 64 = 8^2.
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MATHEMATICA
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Rest@ CoefficientList[Series[x (1 + 12 x - 530 x^2 + 12 x^3 + x^4)/((1 - x) (1 - 38 x + x^2) (1 + 38 x + x^2)), {x, 0, 17}], x] (* Michael De Vlieger, Jun 27 2016 *)
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PROG
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(PARI) Vec(x*(1+12*x-530*x^2+12*x^3+x^4)/((1-x)*(1-38*x+x^2)*(1+38*x+x^2))+ O(x^20))
(PARI) isok(n) = issquare(2*n-1) && issquare(5*n-1); \\ Michel Marcus, Jun 28 2016
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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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STATUS
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approved
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