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A352756
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Positive numbers k such that the centered cube number k^3 + (k+1)^3 is equal to the difference of two positive cubes and to A352755(n).
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8
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3, 46, 197, 528, 1111, 2018, 3321, 5092, 7403, 10326, 13933, 18296, 23487, 29578, 36641, 44748, 53971, 64382, 76053, 89056, 103463, 119346, 136777, 155828, 176571, 199078, 223421, 249672, 277903, 308186, 340593, 375196, 412067, 451278, 492901, 537008, 583671, 632962, 684953, 739716, 797323, 857846, 921357
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OFFSET
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1,1
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COMMENTS
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Numbers B > 0 such that the centered cube number B^3 + (B+1)^3 is equal to the difference of two positive cubes, i.e., A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, with C > D > B > 0, and A > 0, A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = A352755(n), B = a(n) (this sequence), C = A352757(n) and D = A352758(n).
There are infinitely many such numbers a(n) = B 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.
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FORMULA
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a(n) = ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1)/2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 72, with a(1) = 3, a(2) = 46 and a(3) = 197.
a(n) can be extended for negative n such that a(-n) = -a(n+1) - 1.
G.f.: x*(3 + 34*x + 31*x^2 + 4*x^3)/(1 - x)^4. - Stefano Spezia, Apr 08 2022
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EXAMPLE
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a(1) = 3 is a term because 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 46 is a term because 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) - 1)/2 = 197.
a(4) = 3*197 - 3*46 + 3 + 72 = 528.
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MAPLE
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restart; for n to 20 do (1/2)* ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1); end do;
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CROSSREFS
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Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352222, A352223, A352224, A352225, A352755, A352757, A352758, A352759, A352760, A352761, A352762.
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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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