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A352755
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Positive centered cube numbers that can be written as the difference of two positive cubes: a(n) = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1 and n > 0.
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8
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91, 201159, 15407765, 295233841, 2746367559, 16448122691, 73287987409, 264133278045, 811598515091, 2202365761759, 5410166901741, 12249942682409, 25914353312575, 51755729480091, 98389720844009, 179211321358741, 314429627203659, 533744613620855, 879807401606341, 1412624924155809
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OFFSET
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1,1
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COMMENTS
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Numbers A > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, with C > D > B > 0, and A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A352756(n), C = A352757(n) and D = A352758(n).
There are infinitely many such numbers a(n) = A in this sequence.
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LINKS
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A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
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FORMULA
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a(n) = (2*n - 1)*(3*(2*n - 1)^2 + 4)*((2*n - 1)^2*(3*(2*n - 1)^2 + 4)^2 + 3)/4.
a(n) can be extended for negative n such that a(-n) = -a(n+1).
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EXAMPLE
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a(1) = 91 belongs to the sequence because 91 = 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 201159 belongs to the sequence because 201159 = 46^3 + 47^3 = 151^3 - 148^3 and 151 - 148 = 3 = 2*2 - 1.
a(3) = (2*3 - 1)*(3*(2*3 - 1)^2 + 4)*((2*3 - 1)^2*(3*(2*3 - 1)^2 + 4)^2 + 3)/4 = 15407765.
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MAPLE
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restart; for n to 20 do (1/4)*(2*n-1)*(3*(2*n-1)^2+4)*((2*n-1)^2*(3*(2*n-1)^2+4)^2+3) end do;
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CROSSREFS
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Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352221, A352222, A352223, A352224, A352225, A352756, A352757, A352758, A352759, A355751, A355752, A355753.
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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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